Comparing Two Independent Population Proportions Proportions have their base in the binomial probability distribution. As a probability distribution, we know the mean and standard deviation of the distribution. As a binary or categorical sample data set, we lose this knowledge from when we knew the population parameters. We do not know the population mean, µ = n p , or variance, σ ∠ = n p q . We can gather data that we know comes from a binary distribution but not know the specific parameter. It is then that we have moved from probability to inferential statistics. When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: The two independent samples are random samples that are independent. The number of successes is at least five, and the number of failures is at least five, for each of the samples. Growing literature states that the population must be at least ten or even perhaps 20 times the size of the sample. This keeps each population from being over-sampled and causing biased results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance in the sampling. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the two population proportions. Like the case of differences in sample means, we construct a sampling distribution for differences in sample proportions: ( p A ' - p B ' ) where p ' A = X A n A and p ' B = X ' B n B are the sample proportions for the two sets of data in question. X A and X B are the number of successes in each sample group respectively, and n A and n B are the respective sample sizes from the two groups. Again we go the Central Limit theorem to find the distribution of this sampling distribution for the differences in sample proportions. And again we find that this sampling distribution, like the ones past, are normally distributed as proved by the Central Limit Theorem, as seen in . Generally, the null hypothesis allows for the test of a difference of a particular value, 𝛿 0 , just as we did for the case of differences in means. H 0 : p 1 − p 2 = 𝛿 0 H 1 : p 1 − p 2 ≠ 𝛿 0 Most common, however, is the test that the two proportions are the same. That is, H 0 : p A = p B H a : p A ≠ p B To conduct the test, we use a pooled proportion, p c . The pooled proportion is calculated as follows: p c = x A + x B n A + n B The test statistic ( z -score) is: Z c = ( p A ′ − p B ′ ) − δ 0 p c ( 1 − p c ) ( 1 n A + 1 n B ) where δ 0 is the hypothesized differences between the two proportions and p c is the pooled variance from the formula above. A bank has recently acquired a new branch and thus has customers in this new territory. They are interested in the default rate in their new territory. They wish to test the hypothesis that the default rate is different from their current customer base. They sample 200 files in area A, their current customers, and find that 20 have defaulted. In area B, the new customers, another sample of 200 files shows 12 have defaulted on their loans. At a 10% level of significance can we say that the default rates are the same or different? This is a test of proportions. We know this because the underlying random variable is binary, default or not default. Further, we know it is a test of differences in proportions because we have two sample groups, the current customer base and the newly acquired customer base. Let A and B be the subscripts for the two customer groups. Then p A and p B are the two population proportions we wish to test. Random Variable: P′ A – P′ B = difference in the proportions of customers who defaulted in the two groups. H 0 : p A = p B H a : p A ≠ p B The words "is a difference" tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: p c = x A + x B n A + n B = 20 + 12 200 + 200 = 0.08 1 – p c = 0.92 ( p′ A – p′ B ) = 0.04 follows an approximate normal distribution. Estimated proportion for group A: p ′ A = x A n A = 20 200 = 0.1 Estimated proportion for group B: p ′ B = x B n B = 12 200 = 0.06 The estimated difference between the two groups is : p′ A – p′ B = 0.1 – 0.06 = 0.04. Z c = ( P′ A − P′ B ) − δ 0 P c ( 1 − P c ) ( 1 n A + 1 n B ) = 1.47 The calculated test statistic is 1.47 and is not in the tail of the distribution. Make a decision: Since the calculate test statistic is not in the tail of the distribution we cannot reject H 0 . Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference between the proportions of customers who defaulted in the two groups. Try It Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. The p -value is 0.0379, so we can reject the null hypothesis. At the 5% significance level, the data support that there is a difference in the pressure tolerances between the two valves. References Data from Educational Resources , December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994. “State of the States.” Gallup, 2013. Available online at http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive (accessed June 17, 2013). “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/westnile/index.htm (accessed June 17, 2013). Chapter Review Test of two population proportions from independent samples. Random variable: p' A – p' B = difference between the two estimated proportions Distribution: normal distribution Formula Review Pooled Proportion: p c = x A + x B n A + n B Test Statistic ( z -score): Z c = ( p ′ A − p ′ B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) where p A ' and p B ' are the sample proportions, p A and p B are the population proportions, P c is the pooled proportion, and n A and n B are the sample sizes. Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS 1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS 2 had system failures within the first eight hours of operation. OS 2 is believed to be more stable (have fewer crashes) than OS 1 . Is this a test of means or proportions? What is the random variable? P ′ OS1 – P ′ OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS 1 and OS 2 . State the null and alternative hypotheses. What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. Is this a test of means or proportions? proportions State the null and alternative hypotheses. H 0 : _________ H a : _________ Is this a right-tailed, left-tailed, or two-tailed test? How do you know? right-tailed What is the random variable of interest for this test? In words, define the random variable for this test. The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for the Exercise 10.39 . Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. Calculate the test statistic. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Cannot accept the null hypothesis. p -value < alpha At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? Homework A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. We are interested in whether the proportions of conferred physical science degrees for people aged 21 to 24 are the same for White people and Black people in the United States. The number of conferred degrees for White people in a given year is 4,930. Five hundred eighty were aged 21 to 24. The estimate for Black people is 330. Forty were aged 21 to 24. We will let people with conferred physical science degrees be our population. H 0 : P W = P B H a : P W ≠ P B The random variable is the difference in the proportions of White and Black people with conferred degrees in a given year, aged 21 to 24. normal for two proportions test statistic: –0.1944 p -value: 0.8458 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of White and Black people with conferred physical science degrees, aged 21 to 24, are different. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic/Latino students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic/Latino students out of a total of 2,441 students. In general, do you think that the percent of Hispanic/Latino students at the two colleges is basically the same or different? Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The random variable is the difference between the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College. normal for two proportions test statistic: 4.29 p -value: 0.00002 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College are different. Use the following information to answer the next two exercises. Some individuals who were exposed to the COVID-19 virus experienced a loss of taste. In a sample of COVID patients during 2022, there were 629 reported cases of loss of taste out of a total of 1,021 reported cases. In 2021, there were 486 reported cases of loss of taste out of a sample of 712 cases. Is the 2022 proportion of loss of taste more than the 2021 proportion? Using a 1% level of significance, conduct an appropriate hypothesis test. “2022” subscript: 2022 group. “2021” subscript: 2021 group This is: a test of two proportions a test of two independent means a test of a single mean a test of matched pairs An appropriate null hypothesis is: p 2022 ≤ p 2021 p 2022 ≥ p 2021 μ 2022 ≤ μ 2021 p 2022 > p 2021 a Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. Test: two independent sample proportions. Random variable: p ′ 1 - p ′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed Adults aged 18 years old and older from a certain city were randomly selected for a survey on working from home. The researchers wanted to determine if the proportion of women who work from home is less than the proportion of men who work from home. The results are shown in . Test at the 1% level of significance. Number who work at home Sample size Men 42,769 155,525 Women 67,169 248,775 Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a tablet 69 231 Sample size 628 2,309 Test: two independent sample proportions Random variable: p′ 1 − p′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 > p 2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: right-tailed Do not reject the H 0 . Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. A group of friends debated whether more people in their twenties use wearable fitness devices than people in their thirties. They consulted a research study of wearable fitness device use among adults. The results of the survey indicate that of the 973 randomly sampled people in their twenties, 379 use such devices. For people in their thirties, 404 of the 1,304 who were randomly sampled use wearable fitness devices. Test at the 5% level of significance. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Subscripts: 1: men; 2: women H 0 : p 1 ≤ p 2 H a : p 1 > p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. normal for two proportions test statistic: 0.22 p -value: 0.4133 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. Joan Nguyen recently claimed that the proportion of college-age men with at least one pierced ear is as high as the proportion of college-age women. She conducted a survey in her classes. Out of 107 men, 20 had at least one pierced ear. Out of 92 women, 47 had at least one pierced ear. Do you believe that the proportion of men has reached the proportion of women? H 0 : p 1 = p 2 H a : p 1 ≠ p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women that have at least one pierced ear. normal for two proportions test statistic: –4.82 p -value: zero Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different. "To Breakfast or Not to Breakfast?" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what they have achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in , solve our problem. Work hours with breakfast Work hours without breakfast 8 6 7 5 9 5 5 4 9 7 8 7 10 7 7 5 6 6 9 5 H 0 : µ d = 0 H a : µ d > 0 The random variable X d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t 9 test statistic: 4.8963 p -value: 0.0004 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.

Math


Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is X 1 – – X 2 – . The normal distribution has the following format: The standard deviation is: ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 The test statistic ( z -score) is: Z c = ( x – 1 – x – 2 ) – δ 0 ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax . Both populations have a normal distributions. The data are recorded in . Wax Sample mean number of months floor wax lasts Population standard deviation 1 3 0.33 2 2.9 0.36 Does the data indicate that wax 1 is more effective than wax 2 ? Test at a 5% level of significance. This is a test of two independent groups, two population means, population standard deviations known. Random Variable : X – 1 – X – 2 = difference in the mean number of months the competing floor waxes last. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The words "is more effective" says that wax 1 lasts longer than wax 2 , on average. "Longer" is a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula for the test statistic we find the calculated value for the problem. Z c = ( μ 1 - μ 2 ) - δ 0 σ 1 2 n 1 + σ 2 2 n 2 = 0.1 The estimated difference between he two means is : X – 1 – X – 2 = 3 – 2.9 = 0.1 Compare calculated value and critical value and Z α : We mark the calculated value on the graph and find the calculated value is not in the tail therefore we cannot reject the null hypothesis. Make a decision: the calculated value of the test statistic is not in the tail, therefore you cannot reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts. Try It The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines of each type are randomly assigned to be tested. Both populations have normal distributions. shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine Sample mean number of RPM Population standard deviation 1 1,500 50 2 1,600 60 The p -value is almost zero, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1. An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance. This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators Random variable: X – 1 – X – 2 = difference in the mean age of Democratic and Republican U.S. senators. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H a : μ 1 > μ 2 H a : μ 1 - μ 2 > 0 The words "older than" translates as a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Make a decision: The p-value is larger than 5%, therefore we cannot reject the null hypothesis. By calculating the test statistic we would find that the test statistic does not fall in the tail, therefore we cannot reject the null hypothesis. We reach the same conclusion using either method of a making this statistical decision. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. Try It The average age of 10 professors selected randomly in university A is 46.672 with a standard deviation of 8.53. The average age of 10 professors selected randomly in university B is 47.531 with a standard deviation of 7.83. Does the data indicate that university A has older professors than university B, on average? Test at a 5% level of significance. The total sample size is 10 + 10 = 20., which is greater than 10, so we can use the Student’s t -distribution. The random variable is X 1 - X 2 = difference between the mean age of universities A and B. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H 0 : μ 1 > μ 2 H 0 : μ 1 - μ 2 > 0 The phrase “older than” translates as a “>” symbol and goes into H 0 . Therefore, this is a right-tailed test. The distribution is: X 1 - X 2 ~ N 0 , ( 8 . 53 ) 2 10 + ( 7 . 83 ) 2 10 The mean for the normal distribution is 0 since μ 1 ≤ μ 2 and μ 1 - μ 2 ≤ 0 . The p -value is obtained as 0.4258. Since alpha is less than the p -value, do not reject the H 0 . At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of professors in university A is more than the mean age of professors in university B. References Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbullying Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). “State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013). Chapter Review A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics: Random variable: X – 1 − X – 2 = the difference of the means Distribution: normal distribution Formula Review Test Statistic ( z -score): Z c = ( x – 1 − x – 2 ) − δ 0 ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 where: σ 1 and σ 2 are the known population standard deviations. n 1 and n 2 are the sample sizes. x – 1 and x – 2 are the sample means. μ 1 and μ 2 are the population means. Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample mean speed of pitches (mph) Population standard deviation Wesley 86 3 Rodriguez 91 7 What is the random variable? The difference in mean speeds of the fastball pitches of the two pitchers State the null and alternative hypotheses. What is the test statistic? –2.46 At the 1% significance level, what is your conclusion? At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant group Sample mean height of plants (inches) Population standard deviation Food 16 2.5 No food 14 1.5 Is the population standard deviation known or unknown? State the null and alternative hypotheses. Subscripts: 1 = Food, 2 = No Food H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample mean melting temperatures (°F) Population standard deviation Alloy Gamma 800 95 Alloy Zeta 900 105 State the null and alternative hypotheses. Subscripts: 1 = Gamma, 2 = Zeta H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 Is this a right-, left-, or two-tailed test? At the 1% significance level, what is your conclusion? There is sufficient evidence so we cannot accept the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. Homework NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. Subscripts: 1 = boys, 2 = girls H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The random variable is the difference in the mean auto insurance costs for boys and girls. normal test statistic: z = 2.50 p -value: 0.0062 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. normal test statistic: 6.36 p -value: 0 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s score 2 2 3 3 4 2 1 1 2 4 Husband’s score 2 2 1 3 2 1 1 1 2 4 H 0 : µ d = 0 H a : µ d < 0 The random variable X d is the average difference between husband’s and wife’s satisfaction level. t 9 test statistic: t = –1.86 p -value: 0.0479 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis, but run another test. Reason for Decision: p -value < alpha Conclusion: This is a weak test because alpha and the p -value are close. However, there is insufficient evidence to conclude that the mean difference is negative.

Math


Matched or Paired Samples In most cases of economic or business data we have little or no control over the process of how the data are gathered. In this sense the data are not the result of a planned controlled experiment. In some cases, however, we can develop data that are part of a controlled experiment. This situation occurs frequently in quality control situations. Imagine that the production rates of two machines built to the same design, but at different manufacturing plants, are being tested for differences in some production metric such as speed of output or meeting some production specification such as strength of the product. The test is the same in format to what we have been testing, but here we can have matched pairs for which we can test if differences exist. Each observation has its matched pair against which differences are calculated. First, the differences in the metric to be tested between the two lists of observations must be calculated, and this is typically labeled with the letter "d." Then, the average of these matched differences, X – d = ∑ ( x 1 - x 2 ) n is calculated as is its standard deviation, S d . We expect that the standard deviation of the differences of the matched pairs will be smaller than unmatched pairs because presumably fewer differences should exist because of the correlation between the two groups. When using a hypothesis test for matched or paired samples, the following characteristics may be present: Simple random sampling is used. Sample sizes are often small. Two measurements (samples) are drawn from the same pair of individuals or objects. Differences are calculated from the matched or paired samples. The differences from the sample that is used for the hypothesis test. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal, as has been the case using the Central Limit Theorem. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences, that is, the number of pairs not the number of observations. The null and alternative hypotheses for this test are: H 0 : µ d = 0 H a : µ d ≠ 0 The test statistic is: t c = x – d − μ d ( s d n ) A company has developed a training program for its entering employees because management has become concerned with the results of the six-month employee review. They hope that the training program can result in better six-month reviews. Each trainee constitutes a “pair”, the entering score the employee received when first entering the firm and the score given at the six-month review. The difference in the two scores were calculated for each employee and the means for before and after the training program was calculated. The sample mean before the training program was 20.4 and the sample mean after the training program was 23.9. The standard deviation of the differences in the two scores across the 20 employees was 3.8 points. Test at the 10% significance level the null hypothesis that the two population means are equal against the alternative that the training program helps improve the employees’ scores. The first step is to identify this as a two sample case: before the training and after the training. This differentiates this problem from simple one sample issues. Second, we determine that the two samples are "paired." Each observation in the first sample has a paired observation in the second sample. This information tells us that the null and alternative hypotheses should be: H 0 : µ d ≤ 0 H a : µ d > 0 This form reflects the implied claim that the training course improves scores; the test is one-tailed and the claim is in the alternative hypothesis. Because the experiment was conducted as a matched paired sample rather than simply taking scores from people who took the training course those who didn't, we use the matched pair test statistic: Test Statistic: t c = X ¯ d − µ d S d n = ( 23.9 − 20.4 ) − 0 ( 3.8 20 ) = 4.12 In order to solve this equation, the individual scores, pre-training course and post-training course need to be used to calculate the individual differences. These scores are then averaged and the average difference is calculated: X ¯ d = ∑ ( x 1 - x 2 ) n From these differences we can calculate the standard deviation across the individual differences: S d = ∑ d i - X ¯ d 2 n - 1 w h e r e d i = x 1 i - x 2 i We can now compare the calculated value of the test statistic, 4.12, with the critical value. The critical value is a Student's t with degrees of freedom equal to the number of pairs, not observations, minus 1. In this case 20 pairs and at 90% confidence level t a/2 = ±1.729 at df = 20 - 1 = 19. The calculated test statistic is most certainly in the tail of the distribution and thus we cannot accept the null hypothesis that there is no difference from the training program. Evidence seems indicate that the training aids employees in gaining higher scores. Try-It Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 111 135 140 125 Off-hand 105 109 98 111 99 A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in . A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.") After data Before data Difference 6.8 6.6 0.2 2.4 6.5 -4.1 7.4 9 -1.6 8.5 10.3 -1.8 8.1 11.3 -3.2 6.1 8.1 -2 3.4 6.3 -2.9 2 11.6 -9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: x – d = –3.13 and s d = 2.91 Verify these values. Let μ d be the population mean for the differences. We use the subscript d to denote "differences." Random variable: X – d = the mean difference of the sensory measurements H 0 : μ d ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ d is the population mean of the differences.) H a : μ d < 0 The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t 7 . (Notice that the test is for a single population mean.) Calculate the test statistic and look up the critical value using the Student's-t distribution: The calculated value of the test statistic is 3.06 and the critical value of the t -distribution with 7 degrees of freedom at the 5% level of confidence is 1.895 with a one-tailed test. X – d is the random variable for the differences. The sample mean and sample standard deviation of the differences are: x – d = –3.13 s – d = 2.91 Compare the critical value for alpha against the calculated test statistic. The conclusion from using the comparison of the calculated test statistic and the critical value will gives us the result. In this question the calculated test statistic is 3.06 and the critical value is 1.895. The test statistic is clearly in the tail and thus we cannot accept the null hypotheses that there is no difference between the two situations, hypnotized and not hypnotized. Make a decision: Cannot accept the null hypothesis, H 0 . This means that μ d < 0 and there is a statistically significant improvement. Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain. Try It A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level. Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 A college softball coach was interested in whether the college's strength development class increased their players' maximum lift (in pounds). Four players were asked to participate in the study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 241 338 368 Amount of weight lifted after the class 295 252 330 360 The coach wants to know if the strength development class makes the players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. x – d = 21.3, s d = 46.7 Using the difference data, this becomes a test of a single mean. Define the random variable: X – d mean difference in the maximum lift per player. The distribution for the hypothesis test is a student's t with 3 degrees of freedom. H 0 : μ d ≤ 0, H a : μ d > 0 Calculate the test statistic look up the critical value: The calculated value of the test statistic is 0.91. The critical value of the student's t at 5% level of significance and 3 degrees of freedom is 2.353. Decision: If the level of significance is 5%, we cannot reject the null hypothesis, because the calculated value of the test statistic is not in the tail. What is the conclusion? At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. Be sure to note in your conclusion that a sample size of only 4 leaves a degree of freedom of only 3 and sets the critical value very large. In short, sample size so small cannot result in always meaningful conclusions. Try It A new prep class was designed to improve SAT test scores. Four students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in . Are the scores, on average, higher after the class? Test at a 5% level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 1960 1920 2150 Score after class 1920 2160 2200 2100 Chapter Review A hypothesis test for matched or paired samples (t-test) has these characteristics: Test the differences by subtracting one measurement from the other measurement Random Variable: x – d = mean of the differences Distribution: Student’s-t distribution with n – 1 degrees of freedom If the number of differences is small (less than 30), the differences must follow a normal distribution. Two samples are drawn from the same set of objects. Samples are dependent. Formula Review Test Statistic ( t -score): t c = x – d − μ d ( s d n ) where: x – d is the mean of the sample differences. μ d is the mean of the population differences. s d is the sample standard deviation of the differences. n is the sample size. Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in . The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. Installation A B C D E F G H Before 3 6 4 2 5 8 2 6 After 1 5 2 0 1 0 2 2 What is the random variable? the mean difference of the system failures State the null and alternative hypotheses. What conclusion can you draw about the software patch? With a p -value 0.0067, we can cannot accept the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. Subject A B C D E F Before 3 4 3 2 4 5 After 4 5 6 4 5 7 State the null and alternative hypotheses. What is the sample mean difference? What conclusion can you draw about the juggling class? Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 State the null and alternative hypotheses. H 0 : μ d ≥ 0 H a : μ d < 0 What is the test statistic? What is the sample mean difference? What is the conclusion? We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. Homework Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in . Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 140 220 230 110 120 240 220 200 190 180 150 190 200 360 300 280 300 260 240 p -value = 0.1494 At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. A new heart disease prevention drug was tried on a group of 224 patients. Forty-five patients developed heart disease after four years. In a control group of 224 patients, 68 developed heart disease after four years. We want to test whether the method of treatment reduces the proportion of patients that develop heart disease after four years or if the proportions of the treated group and the untreated group stay the same. Let the subscript t = treated patient and ut = untreated patient. The appropriate hypotheses are: H 0 : p t < p ut and H a : p t ≥ p ut H 0 : p t ≤ p ut and H a : p t > p ut H 0 : p t = p ut and H a : p t ≠ p ut H 0 : p t = p ut and H a : p t < p ut An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: x – d = −10.2 s d = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training. The distribution for the test is: t 5 t 6 N (−10.2, 8.4) N(−10.2, 8.4 6 ) A golf instructor is interested in determining if their new technique for improving players’ golf scores is effective. The instructor takes four new students and records their 18-hole scores before learning the technique and then after having taken the class. The instructor conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 The correct decision is: Reject H 0 . Do not reject the H 0 . A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. The group compared the estimates of new female breast cancer cases by southern state in Year 1 and in Year 2. The results are in . Southern states Year 1 Year 2 Alabama 3,450 3,720 Arkansas 2,150 2,280 Florida 15,540 15,710 Georgia 6,970 7,310 Kentucky 3,160 3,300 Louisiana 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas 15,050 14,980 Virginia 6,190 6,280 Test: two matched pairs or paired samples ( t -test) Random variable: X – d Distribution: t 12 H 0 : μ d = 0 H a : μ d > 0 The mean of the differences of new female breast cancer cases in the south between Year 2 and Year 1 is greater than zero. The estimate for new female breast cancer cases in the south is higher in Year 2 than in Year 1. Graph: right-tailed p -value: 0.0004 Decision: Cannot accept H 0 Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in Year 2 than in Year 1. A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in . Test at the 1% level of significance. Cities Hyatt Regency prices in dollars Hilton prices in dollars Atlanta 107 169 Boston 358 289 Chicago 209 299 Dallas 209 198 Denver 167 169 Indianapolis 179 214 Los Angeles 179 169 New York City 625 459 Philadelphia 179 159 Washington, DC 245 239 A politician asked their staff to determine whether the underemployment rate in the northeast decreased year over year. The results are in . Northeastern states Year 1 Year 2 Connecticut 17.3 16.4 Delaware 17.4 13.7 Maine 19.3 16.1 Maryland 16.0 15.5 Massachusetts 17.6 18.2 New Hampshire 15.4 13.5 New Jersey 19.2 18.7 New York 18.5 18.7 Ohio 18.2 18.8 Pennsylvania 16.5 16.9 Rhode Island 20.7 22.4 Vermont 14.7 12.3 West Virginia 15.5 17.3 Test: matched or paired samples ( t -test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8} Random Variable: X – d Distribution: H 0 : μ d = 0 H a : μ d < 0 The mean of the differences of the rate of underemployment in the northeastern states between Year 2 and Year 1 is less than zero. The underemployment rate went down from Year 1 to Year 2. Graph: left-tailed. Decision: Cannot reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from Year 1 to Year 2. Bringing It Together Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. independent group means, population standard deviations and/or variances known independent group means, population standard deviations and/or variances unknown matched or paired samples single mean two proportions single proportion A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. e The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from nine men and 16 women. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. d A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively. According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this. f According to a recent study, U.S. companies have a mean maternity-leave of six weeks. A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. e A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 300 960 920 1010 1100 840 880 1100 1070 1250 1320 860 860 1330 1370 790 770 990 1040 1110 1200 740 850 University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked. f Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown . Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size 41 41 Sample mean 97.5 98.1 Sample standard deviation 17.5 19.2 Two independent means, normal distribution Two independent means, Student’s-t distribution Matched or paired samples, Student’s-t distribution Two population proportions, normal distribution A golf instructor is interested in determining if a new technique for improving players’ golf scores is effective. The instructor takes four (4) new students and records their 18-hole scores before learning the technique and then after having taken the class. The instructor conducts a hypothesis test. The data are as . Player 1 Player 2 Player 3 Player 4 Mean score before class 83 78 93 87 Mean score after class 80 80 86 86 This is: a test of two independent means. a test of two proportions. a test of a single mean. a test of a single proportion. a

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Introduction The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: modification of work “The first few times I used my Asda 'bag for life' I left the receipts in the bottom of the bag” by Pete/ Flickr, Public domain) Have you ever wondered if lottery winning numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example the test of independence, which determines if events are independent, such as in the movie example the test of a single variance, which tests variability, such as in the coffee example

Math


Facts About the Chi-Square Distribution The notation for the chi-square distribution is: X ∼ χ d f 2 where df = degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chi-square probabilities then use df = n - 1. The degrees of freedom for the three major uses are each calculated differently.) For the χ 2 distribution, the population mean is μ = df and the population standard deviation is σ = 2 ( d f ) . The random variable is shown as χ 2 . The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables. χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + ... + ( Z k ) 2 The curve is nonsymmetrical and skewed to the right. There is a different chi-square curve for each df . The test statistic for any test is always greater than or equal to zero. When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ 1,000 2 the mean, μ = df = 1,000 and the standard deviation, σ = 2 ( 1,000 ) = 44.7. Therefore, X ~ N (1,000, 44.7), approximately. The mean, μ , is located just to the right of the peak. References Data from Parade Magazine . “HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011. Chapter Review The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df . For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. Formula Review χ 2 = ( Z 1 ) 2 + ( Z 2 ) 2 + … ( Z df ) 2 chi-square distribution random variable μ χ 2 = df chi-square distribution population mean σ χ 2 = 2 ( d f ) Chi-Square distribution population standard deviation If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? mean = 25 and standard deviation = 7.0711 If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. When does the chi-square curve approximate a normal distribution? when the number of degrees of freedom is greater than 90 Where is μ located on a chi-square curve? Is it more likely the df is 90, 20, or two in the graph? df = 2 Homework Decide whether the following statements are true or false. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. true The standard deviation of the chi-square distribution is twice the mean. The mean and the median of the chi-square distribution are the same if df = 24. false

Math


Test of a Single Variance Thus far our interest has been exclusively on the population parameter μ or it's counterpart in the binomial, p. Surely the mean of a population is the most critical piece of information to have, but in some cases we are interested in the variability of the outcomes of some distribution. In almost all production processes quality is measured not only by how closely the machine matches the target, but also the variability of the process. If one were filling bags with potato chips not only would there be interest in the average weight of the bag, but also how much variation there was in the weights. No one wants to be assured that the average weight is accurate when their bag has no chips. Electricity voltage may meet some average level, but great variability, spikes, can cause serious damage to electrical machines, especially computers. I would not only like to have a high mean grade in my classes, but also low variation about this mean. In short, statistical tests concerning the variance of a distribution have great value and many applications. A test of a single variance assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population variance . The test statistic is: χ c 2 = ( n - 1 ) s 2 σ 0 2 where: n = the total number of observations in the sample data s 2 = sample variance σ 0 2 = hypothesized value of the population variance H 0 : σ 2 = σ 0 2 H a : σ 2 ≠ σ 0 2 You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Even though we are given the population standard deviation, we can set up the test using the population variance as follows. H 0 : σ 2 ≤ 5 2 H a : σ 2 > 5 2 Try It A SCUBA instructor wants to record the collective depths each of his students' dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? H 0 : σ 2 = 3 2 H a : σ 2 < 3 2 With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon. With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers . Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ 2 . Random Variable: The sample standard deviation, s , is the random variable. Let s = standard deviation for the waiting times. H 0 : σ 2 ≥ 7.2 2 H a : σ 2 < 7.2 2 The word "less" tells you this is a left-tailed test. Distribution for the test: χ 24 2 , where: n = the number of customers sampled df = n – 1 = 25 – 1 = 24 Calculate the test statistic: χ c 2 = ( n − 1 ) s 2 σ 2 = ( 25 − 1 ) ( 3.5 ) 2 7.2 2 = 5.67 where n = 25, s = 3.5, and σ = 7.2. The graph of the Chi-square shows the distribution and marks the critical value with 24 degrees of freedom at 95% level of confidence, α = 0.05, 13.85. The critical value of 13.85 came from the Chi squared table which is read very much like the students t table. The difference is that the students t -distribution is symmetrical and the Chi squared distribution is not. At the top of the Chi squared table we see not only the familiar 0.05, 0.10, etc. but also 0.95, 0.975, etc. These are the columns used to find the left hand critical value. The graph also marks the calculated χ 2 test statistic of 5.67. Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion. Make a decision: Because the calculated test statistic is in the tail we cannot accept H 0 . This means that you reject σ 2 ≥ 7.2 2 . In other words, you do not think the variation in waiting times is 7.2 minutes or more; you think the variation in waiting times is less. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. Try It The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p -value, and draw a conclusion. Test at the 1% significance level. Professor Holmes has a weakness for cream filled donuts, but they believe that some bakeries are not properly filling the donuts. A sample of 24 donuts reveals a mean amount of filling equal to 0.04 cups, and the sample standard deviation is 0.11 cups. Professor Holmes has an interest in the average quantity of filling, of course, but he is particularly distressed if one donut is radically different from another. Professor Holmes does not like surprises. Test at 95% the null hypothesis that the population variance of donut filling is significantly different from the average amount of filling 0.04 cups. This is clearly a problem dealing with variances. In this case we are testing a single sample rather than comparing two samples from different populations. The null and alternative hypotheses are thus: H 0 : σ 2 = 0.04 H a : σ 2 ≠ 0.04 The test is set up as a two-tailed test because Professor Holmes has shown concern with too much variation in filling as well as too little: their dislike of a surprise is a significant difference greater than or less than the expected average of 0.04 cups. Conduct the hypothesis test with n = 24 , s 2 = 0 . 11 , and σ 2 = 0 . 04 . The test statistic is calculated to be: χ c 2 = ( n − 1 ) s 2 σ o 2 = ( 24 − 1 ) 0.11 0.04 = 63.25 The calculated χ 2 test statistic, 63.25, is in the tail. Therefore, at a 0.05 level of significance, we cannot accept the null hypothesis that the variance in the donut filling is equal to 0.04 cups. It seems that Professor Holmes is destined to meet disappointment with each bite. Try It The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. At a certain point in time, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the distribution and mark the area associated with the level of confidence, and draw a conclusion. Test at the 1% significance level. H 0 : σ 2 = 12.2 2 H a : σ 2 > 12.2 2 df = 14 chi 2 test statistic = 16.39 The p -value is 0.2902, so we decline to reject the null hypothesis. There is not enough evidence to suggest that the variance is greater than 12.2 2 . In 2nd DISTR , use7: χ2cdf . The syntax is (lower, upper, df) for the parameter list. χ2cdf(16.39,10^99,14) . The p -value = 0.2902. References “AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013). Data from the World Bank, June 5, 2012. Chapter Review To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation). Formula Review χ 2 = ( n − 1 ) s 2 σ 0 2 Test of a single variance statistic where: n : sample size s : sample standard deviation σ 0 : hypothesized value of the population standard deviation df = n – 1 Degrees of freedom Test of a Single Variance Use the test to determine variation. The degrees of freedom is the number of samples – 1. The test statistic is ( n – 1 ) s 2 σ 0 2 , where n = sample size, s 2 = sample variance, and σ 2 = population variance. The test may be left-, right-, or two-tailed. Use the following information to answer the next three exercises: An archer’s standard deviation for hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less. What type of test should be used? a test of a single variance State the null and alternative hypotheses. Is this a right-tailed, left-tailed, or two-tailed test? a left-tailed test Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. What type of test should be used? State the null and alternative hypotheses. H 0 : σ 2 = 0.81 2 ; H a : σ 2 > 0.81 2 df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. What type of test should be used? a test of a single variance What is the test statistic? What can you conclude at the 5% significance level? Homework Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. Is the traveler disputing the claim about the average or about the variance? A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 225 Is this a right-tailed, left-tailed, or two-tailed test? H 0 : __________ H 0 : σ 2 ≤ 150 df = ________ chi-square test statistic = ________ 36 Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the area associated with the level of confidence. Answers may vary. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________ How did you know to test the variance instead of the mean? The claim is that the variance is no more than 150 minutes. If an additional test were done on the claim of the average delay, which distribution would you use? If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? a Student's t - or normal distribution A plant manager is concerned their equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. H 0 : σ = 15 H a : σ > 15 df = 42 chi-square with df = 42 test statistic = 26.88 Answers may vary. Alpha = 0.05 Decision: Cannot reject null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. H 0 : σ ≤ 3 H a : σ > 3 df = 17 chi-square distribution with df = 17 test statistic = 28.73 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in . Does the students’ survey indicate that the standard deviation is greater than 0.75? # of births Frequency 0 5 1 30 2 10 3 5 According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. They count the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11 H 0 : σ = 2 H a : σ ≠ 2 df = 14 chi-square distribution with df = 14 chi-square test statistic = 5.2094 Answers may vary. Alpha = 0.05 Decision: Cannot accept the null hypothesis Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2. The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. The manager randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis? The sample standard deviation is $34.29. H 0 : σ 2 = 25 2 H a : σ 2 > 25 2 df = n – 1 = 7. test statistic: x 2 = x 7 2 = ( n – 1 ) s 2 25 2 = ( 8 – 1 ) ( 34.29 ) 2 25 2 = 13.169 ; Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution. Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test. (a) at the 5% significance level (b) at the 1% significance level Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172;

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