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. What is one way to accomplish that? e. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. 124. Refer to Exercise 8.123. Another question in the poll was “[How much are] you worried about the quality of educatio... |
Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. a. b. c. d. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) This content is available for free... |
. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. a. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. b. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of th... |
Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/ 2000growthchart-us.pdf (accessed July 2, 2013). 458 CHAPTER 8 | CONFIDENCE INTER... |
ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013). “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/ finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013). 8.3 A Po... |
news/299-americans-neither-worried-norprepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). “52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/ 52_say_big_time... |
students is between 24.52 and 36.28. 37 The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean. ¯ 39 X is the number of hours a patient waits in the emergency room before being cal... |
the sample mean. 75 P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. 77 CI: (0.62735, 0.67265) EBM: 0.02265 79 The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class. 81 a. x = 64 b. n = 80 c. p′ = 0.8 83 p 85 P′ ~ N⎛ ⎝... |
i. (22.228, 24.972) ii. Figure 8.18 iii. EBM = 1.372 e. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size. f. The confidence level would increase as a result of a larger interval. Smaller sa... |
114, $850,632). 8. Notice the small difference between the two solutions–these differences are simply due to rounding error in the hand calculations. d. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. 103 Use the f... |
= 2.150 EBM = t α 2 521, 130.41 30 = 0.02t α 2 = t0.02 s n ⎛ ⎝ - EBM = $251,854.23 - $204,561.66 = $47,292.57 x¯ + EBM = $251,854.23+ $204,561.66 = $456,415.89 We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $45... |
the president is doing an acceptable job. b. N⎛ ⎝0.61, (0.61)(0.39) 1200 ⎞ ⎠ c. i. CI: (0.59, 0.63) ii. Check student’s solution iii. EBM: 0.02 121 a. i. ii. (0.72, 0.82) (0.65, 0.76) iii. (0.60, 0.72) b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.... |
, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. c. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125 z α 2 = 1.150. (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) EBP = (1.150) 0.52(0.48) ... |
deviation known. • Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown. • Conduct and interpret hypothesis tests for a single population proportion. One job of a statistician is to make statistical inferences about populations based on samples taken from the popul... |
we conclude when we reject H0. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decisi... |
hypotheses. a. H0: μ __ 66 b. Ha: μ __ 66 Example 9.3 We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are: H0: μ ≥ 5 Ha: μ < 5 9.3 We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alt... |
is to reject H0 when H0 is true (incorrect decision known as a Type I error). 3. The decision is not to reject H0 when, in fact, H0 is false (incorrect decision known as a Type II error). 4. The decision is to reject H0 when H0 is false (correct decision whose probability is called the Power of the Test). Each of the ... |
The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead. α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P(Type I error). β = prob... |
coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequenc... |
mean, the distribution for the test is for means: ¯ X ~ N⎛ ⎝µ X, σ X n ⎞ ⎠ or td f The population parameter is μ. The estimated value (point estimate) for μ is x¯, the sample mean. If you are testing a single population proportion, the distribution for the test is for proportions or percentages: This content is availa... |
= p and σ = pq n. Remember that q = 1 – p. 9.4 | Rare Events, the Sample, Decision and Conclusion Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when work... |
more clearly. 476 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Example 9.9 Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The ... |
of the baker's loaves of bread is greater than 15 cm. 9.9 A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5. H0: μ ≤ 12 Ha: μ > 12 The p-value is 0.0013 Draw a graph that shows the p-value. This content is... |
. Reject the null hypothesis when ______________________________________. The results of the sample data _____________________________________. Do not reject the null when hypothesis when __________________________________________. The results of the sample data ____________________________________________. Solution 9.... |
decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules. The following examples illustrate a left-, right-, and two-tailed test. Example 9.11 Ho: μ = 5, Ha: μ < 5 Test of a single population mean. Ha tells you the test is left-tailed. The picture of... |
16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Solution 9.14 Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean. H0: μ = 16.43 Ha: μ < 16.43 For Jeffrey to swim faster, his time wi... |
you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles. Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time ... |
Do not reject the null hypothesis when the null hypothesis is false.) 9.14 The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances fo... |
score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two proba... |
σ = 55 pounds (Always use σ if you know it.) We assume μ = 275 pounds unless our data shows us otherwise. Calculate the p-value using the normal distribution for a mean and using the sample mean as input (see Appendix G for using the data as input): p-value = P( x¯ > 286.2) = 0.1323. Interpretation of the p-value: If ... |
is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution. Solution 9.16 Set up the hypot... |
the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not o... |
. Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion. P′ ~ N⎛ ⎝p, p ⋅ q n ⎞ ⎠ Therefore, P′ ~ N⎛ ⎝0.5, 0.5 ⋅ 0.5 100 ⎞ ⎠ where p = 0.50, q = 1−p = 0.50, and n = 100 Calculate the p-value using the ... |
Prop not equals.5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned o... |
⎞ ⎠ ⎝0.30, a. The value that helps determine the p-value is p′. Calculate p′. Solution 9.18 a. p′ = x n where x is the number of successes and n is the total number in the sample. x = 43, n = 150 p′ = 43 150 b. What is a success for this problem? Solution 9.18 b. A success is having three cell phones in a household. c... |
fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! ... |
< p-value Table 9.3 Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%. Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label t... |
OTHESIS TESTING WITH ONE SAMPLE 4. State the Conclusions: Since the p-value* (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one. Example 9.21 In a study of 420,019 cell phone use... |
, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically sig... |
complete sentence. Language Survey About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is di... |
hypothesis. Binomial Distribution a discrete random variable (RV) that arises from Bernoulli trials. There are a fixed number, n, of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the sa... |
hypothesis testing, the Level of Significance is called the preconceived α or the preset α. Normal Distribution a continuous random variable (RV) with pdf f (x) = −(x − µ)2 2σ 2 1 σ 2π e, where μ is the mean of the distribution, and σ is the standard deviation, notation: X ~ N(μ, σ). If μ = 0 and σ = 1, the RV is call... |
, or <). 3. If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. 4. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolut... |
p-value, reject the null hypothesis 2. α ≤ p-value, do not reject the null hypothesis 9.5 Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: 1. Determine H0 and Ha. Remember, they are contradictory. This content is availabl... |
H0 and Ha are contradictory. greater than or equal to (≥) less than or equal to (≤) If Ho has: then Ha has: equal (=) not equal (≠) or greater than (>) or less than (<) Table 9.4 If α ≤ p-value, then do not reject H0. If α > p-value, then reject H0. α is preconceived. Its value is set before the hypothesis test starts... |
length of time in jail from the survey was 3 years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of jail time has increased, what would the null and alternative hypotheses ... |
. What is the probability of a Type II error? 18. A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H0, is: the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. 19. A microbiologist is testing a water sample for E-coli. Suppose the nu... |
of a single population mean using a Student’s t-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? 29. You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq? 30. You are performing a hypothesi... |
.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the pop... |
9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? 50. Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? 51. Assume H0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? 52. Draw the general graph of a left-tailed test. 53.... |
parameter (μ or p). a. The mean number of years Americans work before retiring is 34. b. At most 60% of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. Twenty-nine percent of high school seniors get drunk each month. e. Few... |
: μ = 4.5, Ha: μ > 4.5 9.2 Outcomes and the Type I and Type II Errors 66. State the Type I and Type II errors in complete sentences given the following statements. a. The mean number of years Americans work before retiring is 34. b. At most 60% of Americans vote in presidential elections. c. The mean starting salary fo... |
at least 20%. less than 20%, when in fact, it is less than 20%. 70. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation ... |
Sample, Decision and Conclusion 73. The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depres... |
tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim? 75. From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of arou... |
percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more ... |
for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55. 86. Use the “Lap time” data for Lap 4 (see Appendix C) to test the claim that Terri finishes Lap 4, on... |
hundred one pups: Dalmatians that fetched the most shekels. They asked Mr. Spreckles to name An average spot count he'd claim To bring in big bucks. Said Spreckles, “Well, shucks, It's for one hundred one that I aim.” Said an amateur statistician Who wanted to help with this mission. “Twenty-one for the sample Standar... |
89 • 8 stores @ 0.75. I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starvi... |
end, my Lord? HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no. (Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.) Act II (The n... |
how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right? 95. "Japanese Girls’ Names" by Kumi Furuichi It used to be ... |
, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko. 96. "Phillip’s Wish," by Suzanne Osorio My nephew likes to play Chasing the girls makes... |
teacher, “Sleep no more.” “In every class you fall asleep,” The teacher said, his voice was deep. “So a tally I’ve begun to keep Of every class you nap and snore. The percentage being forty-four.” “My dear teacher I must confess, While sleeping is what I do best. The percentage, I think, must be less, A percentage les... |
surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. 102. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher tha... |
C students who attended the midnight showing of Harry Potter is more than 20%. c. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%. d. There is insufficient evidence to conclude that the percent of EVC students who attended the ... |
% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? 109. Driver error can be listed as the cause of approximately 54% of ... |
on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals 3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1 At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year? 115. According to the N.Y. Times Almanac the mean famil... |
alecheleague.org/Law/BAFeb01.html. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013). Data from the Bureau of Labor Statistics. Available online at http://www.bls.g... |
/frequency-of-sexual-assault (accessed June 27, 2013). SOLUTIONS 1 The random variable is the mean Internet speed in Megabits per second. 3 The random variable is the mean number of children an American family has. 5 The random variable is the proportion of people picked at random in Times Square visiting the city. 7 a... |
49 This is a left-tailed test. 51 This is a two-tailed test. 53 516 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.25 55 a right-tailed test 57 a left-tailed test 59 This is a left-tailed test. 61 This is a two-tailed test. 62 a. H0: μ = 34; Ha: μ ≠ 34 b. H0: p ≤ 0.60; Ha: p > 0.60 c. H0: μ ≥ 100,000; Ha: μ <... |
Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do. f. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number ... |
0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. i. (43,537, 49,463) 76 a. H0: μ = $1.00 b. Ha: μ ≠ $1.00 c. Let X ¯ = the average cost of a daily newspaper. d. normal distribution e. z = –0.866 f. p-value = 0.3865 g. Check student’s solution... |
is (0.411, 0.648) 82 a. H0: μ = 4 b. Ha: μ ≠ 4 ¯ c. Let X the average I.Q. of a set of brown trout. d. e. two-tailed Student's t-test t = 1.95 f. p-value = 0.076 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Concl... |
H0: p = 0.60 b. Ha: p < 0.60 c. Let P′ = the proportion of family members who shed tears at a reunion. d. normal for a single proportion e. –1.71 f. 0.0438 g. Check student’s solution. h. i. alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5% signif... |
solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is mor... |
the proportion of schoolmates who fear public speaking. d. normal for a single proportion e. –1.01 f. p-value = 0.1563 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient eviden... |
≠ 0.488 b. p-value = 0.0114 c. alpha = 0.05 d. Reject the null hypothesis. e. At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks. f. The survey does not appear to be accurate. 110 a. H0: p = 0.517 Ha: p ≠ 0.517 b. p-value = 0.9203. c. alpha = 0.05. d. Do not reject ... |
reject the null hypothesis. e. At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average. f. The student academic group’s claim appears to be correct. 524 CHAPTER 9 | HYPOTHESIS TESTING WITH ONE SAMPLE This content is available for free ... |
to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calcu... |
ESIS TESTING WITH TWO SAMPLES 527 The standard error is: (s1 )2 n1 (s2 )2 n2 + The test statistic (t-score) is calculated as follows: ( x¯ 1 – x¯ 2 ) – (µ 1 – µ 2 ) (s2 )2 n2 + (s1 )2 n1 where: • s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively. • σ1 and σ1 are the unknown population ... |
sports each day? Test at the 5% level of significance. Solution 10.1 The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two popula... |
sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per... |
? Solution 10.2 f. Figure 10.3 right g. What is the p-value? Solution 10.2 g. 0.1928 h. Do you reject or not reject the null hypothesis? Solution 10.2 h. Do not reject. i. Conclusion: Solution 10.2 i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who ... |
69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Table 10.4 Face-to-face Class Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions: a. Is ... |
Figure 10.4 h. Reject the null hypothesis i. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean ... |
Solution 10.5 d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference. 10.5 Weighted alpha is a measure of risk-adjust... |
you reject or not reject the null hypothesis? i. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. j. Calculate Cohen’s d and interpret it. 10.2 | Two Population Means with Known Standard Deviations Even though this situation is not likely (kn... |
test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is⎛ ⎝0, 0.332 20 + 0.362 20 ⎞ ⎠ Since μ1 ≤ μ2 then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p-value using the normal distribution: p-value = 0.1799 Graph: Figure 10..... |
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. Solution 10.7 This is a test of two independent groups, two population means. The population standard deviations are unk... |
The number of successes is at least five, and the number of failures is at least five, for each of the samples. 3. Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two propo... |
A + nB = 20 + 12 200 + 200 = 0.08 1 – pc = 0.92 P′ A – P′B ~ N ⎡ ⎣0, (0.08)(0.92)( 1 200 + 1 200 ⎤ ) ⎦ 538 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. Estimated proportion for group A: p′ A... |
. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING ... |
PLES Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is P = 0.1045 and the test statistic is z = -1.256. Example 10.10 Resear... |
IS TESTING WITH TWO SAMPLES 541 Figure 10.9 Decision: Since α > p-value, reject the H0. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans. TI-83+ and TI-84: Press STAT. Arrow ... |
the sample mean of differences is approximately normal. 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, μd, is then tested using a Student's-t test for a single population mean with ... |
/content/col11562/1.16 CHAPTER 10 | HYPOTHESIS TESTING WITH TWO SAMPLES 543 H0: μd ≥ 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. μd is the population mean of the differences.) Ha: μd < 0 The alternative hypothe... |
Arrow down to μ: and arrow over to < µ 0. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER. 10.11 A study was conducted to investigate how effective a new diet was in lower... |
IS TESTING WITH TWO SAMPLES 545 ¯ Define the random variable: X d mean difference in the maximum lift per player. The distribution for the hypothesis test is t3. H0: μd ≤ 0, Ha: μd > 0 Graph: Figure 10.11 Calculate the p-value: The p-value is 0.2150 Decision: If the level of significance is 5%, the decision is not to r... |
the hands. Distribution for the hypothesis test: t6 H0: μd = 0 Ha: μd ≠ 0 Graph: Figure 10.12 Calculate the p-value: The p-value is 0.0716 (using the data directly). (test statistic = 2.18. p-value = 0.0719 using ⎛ ⎝ x¯ d = 3.71, sd = 4.5. ⎞ ⎠ Decision: Assume α = 0.05. Since α < p-value, Do not reject H0. Conclusion:... |
random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.13 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Wr... |
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