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A study examined the average pay for men and women entering the workforce as doctors for 21 different positions. If each gender was equally paid, then we would expect about half of those positions to have men paid more than women and women would be paid more than men in the other half of positions. Write appropriate hy... | In effect, we're checking whether men are paid more than women (or vice-versa), and we'd expect these outcomes with either chance under the null hypothesis: H0: p = 0.5; HA: p ≠ 0.5 We'll use p to represent the fraction of cases where men are paid more than women. |
A study examined the average pay for men and women entering the workforce as doctors for 21 different positions. Men were, on average, paid more in 19 of those 21 positions. Supposing these 21 positions represent a simple random sample, complete a hypothesis test based on men and women being equally paid. | In effect, we're checking whether men are paid more than women (or vice-versa), and we'd expect these outcomes with either chance under the null hypothesis: H0: p = 0.5; HA: p ≠ 0.5 We'll use p to represent the fraction of cases where men are paid more than women. There isn't a good way to check independence here sinc... |
Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since n ≥ 30. | False. For the distribution of p^ to be approximately normal, we need to have at least 10 successes and 10 failures in our sample (on the average). |
Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. | True. The success-failure condition is not satisfied np = 50 x 0.08 = 4 and n(1 x p) = 50 x 0.92 = 46; therefore we know that the distribution of ^p is not approximately normal. In most samples we would expect p^ to be close to 0.08, the true population proportion. While p^ can be as high as 1 (though we would expect t... |
Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning A random sample of 125 college students where 12% are vegetarians would be considered unusual. | False. Standard error of p^ in samples with n = 125 can be calculated as: SEp^ = sqrt(0.08 x 0.92 / 125) = 0.0243 A p^ of 0.12 is only (0.12 – 0.08) / 0.0243 = 1.65 standard errors away from the mean, which would not be considered unusual. |
Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning A random sample of 250 college students where 12% are vegetarians would be considered unusual. | True. Standard error of p^ in samples with n = 125 can be calculated as: SEp^ = sqrt(0.08 x 0.92 / 250) = 0.0172 A p^ of 0.12 is 0: (12 x 0:08) / 0.0172 = 2.32 standard errors away from the mean, which might be considered unusual. |
Suppose that 8% of college students are vegetarians. Determine if the following statement is true or false, and explain your reasoning The standard error would be reduced by one-half if we increased the sample size from 125 to 250. | False. Since n appears under the square root sign in the formula for the standard error, increasing the sample size from 125 to 250 would decrease the standard error of the sample proportion only by a factor of sqrt(2). |
About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who think they can achieve the American dream in samples of size 20 is left skewed. | True. The success-failure condition is not satisfied np = 20 x 0.77 = 15.4 and n(1 x p) = 20 x 0.23 = 4.6; therefore we know that the distribution of ^p is not approximately normal. In most samples we would expect p^ to be close to 0.77, the true population proportion. While p^ can be as low as 0 (though we would expec... |
About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n ≥ 3... | False. Unlike with means, for the sampling distribution of proportions to be approximately normal, we need to have at least 10 successes and 10 failures in our sample. We do not use n ≥ 30 as a condition to check for the normality of the distribution of p^. |
About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: A random sample of 60 young Americans where 85% think they can achieve the American dream would be considered unusual. | False. Standard error of p^ in samples with n = 60 can be calculated as: SEp^ = sqrt( 0.77 x 0.23 / 60) = 0.384. A p^ of 0.85 is only Z = (0.85-0.77) / 0.0384 = 2.08 standard errors away from the mean, which would be considered unusual. |
About 77% of young adults think they can achieve the American dream. Determine if the following statement is true or false, and explain your reasoning: A random sample of 120 young Americans where 85% think they can achieve the American dream would be considered unusual. | True. Standard error of p^ in samples with n = 120 can be calculated as: SEp^ = sqrt( 0.77 x 0.23 / 120) = 0.046. A p^ of 0.85 is Z = (0.85 – 0.77) / 0.046 = 1.73 standard errors away from the mean, which would not be considered unusual. |
Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 30 is left skewed. | True. The success-failure condition is not satisfied np = 30 x 0.90 = 27 and n(1 - p) = 30 x 0.10 = 3; therefore we know that the distribution of ^p is not nearly normal. In most samples we would expect p^ to be close to 0.90, the true population proportion. While p^ can be as low as 0 (thought we would expect this to ... |
Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. | True. Since n appears in a square root for SE, using a sample size that is 4 times as large will reduce the SE by half. |
Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 140 is approximately normal. | True. The success-failure condition is satisfied np = 140 x 0.90 = 126 and n(1 - p) = 140 x 0.10 = 14; therefore the distribution of p^ is nearly normal. |
Suppose that 90% of orange tabby cats are male. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of random samples of size 280 is approximately normal. | True. The success-failure condition is satisfied np = 280 x 0.90 = 252 and n(1 - p) = 70 x 0.10 = 28; therefore the distribution of p^ is nearly normal. |
About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random s... | True. The success-failure condition is not satisfied np = 12 x 0.25 = 4 and n(1 - p) = 12 x 0.75 = 8; therefore we know that the distribution of p^ is not approximately normal. In most samples we would expect p^ to be close to 0.25, the true population proportion. While p^ can be as high as 1 (though we would expect th... |
About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: In order for the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slum... | True. The minimum sample size that yields at least 10 successes and 10 failures can be calculated as n = max(10 / 0.25, 10 / 0.75) = max(40, 13.3) = 40. We need a sample of at least n = 40 to meet the success failure condition. |
About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: A random sample of 50 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusu... | False. Standard error of ^p in samples with n = 50 can be calculated as: SEp^ = sqrt(0.25 x 0.75 / 50) = 0.0612. A p^ of 0.20 is only Z = (0.20 – 0.25) / 0.0612 = -0.82 standard errors away from the mean, which would not be considered unusual. |
About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: A random sample of 150 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unus... | False. Standard error of p^ in samples with n = 150 can be calculated as: SE^p = sqrt(0.25 x 0.75 / 150) = 0.0354 A p^ of 0.20 is Z = (0.20 – 0.25) / 0.0354 = -1.41 standard errors away from the mean, which would not be considered unusual. |
About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statement is true or false, and explain your reasoning: Tripling the sample size will reduce the standard error of the sample proportion by one-third. | False. Since n appears under the square root sign in the formula for the standard error, doubling the sample size would decrease the standard error of the sample proportion only by a factor of sqrt(3). |
The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin o... | False. A confidence interval is constructed to estimate the population proportion, not the sample proportion. |
The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin o... | True. This is the correct interpretation of the confidence interval, which can be calculated as 0.82 ± 0.02 = (0.80; 0.84). |
The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin o... | True. This is the correct interpretation of the confidence level. |
The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin o... | True. Since the sample size appears under the square root sign in calculation of the standard error, in order to halve the margin of error we would need to quadruple the sample size. |
The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin o... | True. The confidence interval lies entirely above 50%. |
The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level. Ve... | With a random sample from < 10% of the population, independence is satisfied. The success- failure condition is also satisfied. Hence, the margin of error can be calculated as follows: ME = 1.96 x sqrt(0.66 x 0.34 / 1018) = 0.029 ≈ 3% |
The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level. Ba... | A 95% confidence interval for the proportion of adults who think that licensed drivers should be required to re-take their road test once they reach 65 years of age can be calculated as 0.66 ± 0.03 = (0.63; 0.69). Since two thirds (roughly 67%) is contained in the interval we wouldn't reject a null hypothesis where p =... |
A local news outlet reported that 56% of 600 randomly sampled Kansas residents planned to set off fireworks on July 4th. Determine the margin of error for the 56% point estimate using a 95% confidence level. | With a random sample, independence is satisfied. The success-failure condition is also satisfied. Hence, the margin of error can be calculated as follows: ME = 1.96 x sqrt(0.56 x 0.44 / 600) = 0.0397 ≈ 4% |
Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Describe the population parameter of interest. What is the value of the point estimate of ... | The population parameter of interest is the proportion of all Greeks who would rate their lives poorly enough to be considered “suffering", p. The point estimate for this parameter is the proportion of Greeks in this sample who would rate their lives as such, p^ = 0:25. |
Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Check if the conditions required for constructing a confidence interval based on these dat... | 1. Independence: The sample is random, and 1,000 < 10% of all Greeks, therefore the life rating of one Greek in this sample is independent of another. 2. Success-failure: 1,000 x 0.25 = 250 > 10 and 1,000 x 0.75 = 750 > 10. Since the observations are independent and the success-failure condition is met, p^ is expected ... |
Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Construct a 95% confidence interval for the proportion of Greeks who are “suffering”. | A 95% confidence interval can be calculated as follows: 0.25 ± 1.96 sqrt( 0.25 x 0.75 / 1000) = (0.22312, 0.2769) We are 95% confident that approximately 22% to 28% of Greeks would rate their lives poorly enough to be considered “suffering". |
Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Without doing any calculations, describe what would happen to the confidence interval if w... | Increasing the confidence level would increase the margin of error hence widen the interval. |
Greece has faced a severe economic crisis since the end of 2009. A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that 25% of them said they would rate their lives poorly enough to be considered “suffering”. Without doing any calculations, describe what would happen to the confidence interval if w... | Increasing the sample size would decrease the margin of error hence make the interval narrower. |
A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Is this sample a representative sample from the population of all high school seniors in the US? Expla... | No. The sample only represents students who took the SAT, and this was also an online survey. |
A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Let’s suppose the conditions for inference are met. Even if your answer to part (a) indicated that thi... | A 90% confidence interval can be calculated as follows: 0.55 ± 1.65 x sqrt(0.55 x 0.45 / 1509) = (0.5289, 0.5711) We are 95% confident that 53% to 57% of high school seniors are fairly certain that they will participate in a study abroad program in college. |
A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. What does “90% confidence” mean? | 90% of random samples of 1,509 high school seniors would produce a 90% confidence interval that includes the true proportion of high school seniors who took the SAT are fairly and who certain that they will participate in a study abroad program in college. |
A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college. Based on this interval, would it be appropriate to claim that the majority of high school seniors are ... | Yes. The interval lies entirely above 50%. Therefore, at 90% confidence level, it would be appropriate to claim the majority of high school seniors who took the SAT who are fairly certain they will participate in a study abroad program in college. |
The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. Is 61% a sample statistic or a population parameter? Explain. | 61% is a sample statistic, it's the observed sample proportion. |
The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. Construct a 95% confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of t... | A 95% confidence interval can be calculated as follows: 0.61 ± 1.96 x sqrt(0.61(1 – 0.61) / 1578) = (0.586, 0.634) We are 95% confident that approximately 58.6% to 63.4% of Americans think marijuana should be legalized. |
The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. A critic points out that this 95% confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a go... | 1. Independence: The sample is random, and comprises less than 10% of the American population, therefore we can assume that the individuals in this sample are independent of each other 2. Success-failure: The number of successes (people who said marijuana should be legalized: 1578 x 0.61 = 962.58) and failures (people ... |
The General Social Survey asked 1,578 US residents: “Do you think the use of marijuana should be made legal, or not?” 61% of the respondents said it should be made legal. A news piece on this survey’s findings states, “Majority of Americans think marijuana should be legalized.” Based on your confidence interval, is thi... | Yes, the interval is above 50%, suggesting, with 95% confidence, that the true population proportion of Americans who think marijuana should be legalized is greater than 50%. |
A Kaiser Family Foundation poll for US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. A political pundit on TV claims that a majority of Independents support a N... | (i) Let's prepare for running a hypothesis test. We want to check for a majority (or minority), so we use the following hypotheses: H0: p = 0.5 HA: p ≠ 0.5 We have a sample proportion of p^ = 0.55 and a sample size of n = 617 independents. (ii) Next, we check whether the conditions are met to proceed. Since this is a r... |
A Kaiser Family Foundation poll for US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan”. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. Would you expect a confidence interval for the proportion of Independents w... | No. Generally we expect a hypothesis test and a confidence interval to align, so we would expect the confidence interval to show a range of plausible values entirely above 0.5. However, if the confidence level is misaligned (e.g. a 99% confidence level and α = 0.05 significance level), then this is no longer generally ... |
Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school. A newspaper article states that only a minority of the Americans who decide not to go to college do so... | The hypotheses are as follows: H0: p = 0:5 (50% of Americas who decide not to go to college because they cannot afford it do so because they cannot afford it) HA: p < 0:5 (Less than 50% of Americas who decide not to go to college because they cannot afford it do so because they cannot afford it) Before calculating the ... |
Some people claim that they can tell the difference between a diet soda and a regular soda in the first sip. A researcher wanting to test this claim randomly sampled 80 such people. He then filled 80 plain white cups with soda, half diet and half regular through random assignment, and asked each person to take one sip ... | The hypotheses are as follows: H0: p = 0:5 (Results are equivalent to randomly guessing) HA: p ≠ 0:5 (Results are different than just randomly guessing) Before conducting the hypothesis test, we must first check that the conditions for inference are satisfied. 1. Independence: The sample is random, therefore whether or... |
Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it. Calculate a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. | We have previously confirmed that the independence condition is satisfied. We need to recheck the success-failure condition using the sample proportion: 331 x 0.48 = 158.88 > 10 and 331 x 0.52 = 172.12 > 10. An 80% confidence interval can be calculated as follows: 0.48 ± 1.65 x sqrt(0.48 x 0.52 / 331) = (0.435, 0.525) ... |
Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it. Suppose we wanted the margin of error for the 90% confidence level to be about 1.5%. How large of a survey would you recommend? | We are asked to solve for the sample size required to achieve a 1.5% margin of error for a 90% confidence interval and the point estimate is p^ = 0:48. 0.01 ≥ 1.65 x sqrt(0.48 x 0.52 / n) n ≥ 3020.16 ≈ 3121 The sample size n should be at least 3,121. |
Exercise 6.11 presents the results of a poll evaluating support for a generic “National Health Plan” in the US in 2019, reporting that 55% of Independents are supportive. If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size? | Since a sample proportion (p^ = 0.55) is available, we use this for the sample size calculations. The margin of error for a 90% confidence interval is 1.65 x SE We want this to be less than 0.01, where we use p^ in place of p: 1.65 x sqrt( 0.55(1 – 0.55) / n) ≤ 0.01 From this, we get that n must be at least 6739. |
As discussed in Exercise 6.10, the General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey? | We are asked to solve for the sample size required to achieve a 2% margin of error for a 95% confidence interval and the point estimate is p^ = 0.61. 1.96 x sqrt( 0.61(1 – 0.61) / n) ≤ 0.02 n ≥ 2284.792. The sample size n should be at least 2,285. |
A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the followin... | False. Since (pmale - pfemale) is positive, the proportion of males whose favorite color is black is higher than the proportion of females. |
A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the followin... | True. |
A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the followin... | True. |
A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the followin... | True. |
A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A 95% confidence interval for the difference between the proportions of males and females whose favorite color is black (pmale − pfemale) was calculated to be (0.02, 0.06). Based on this information, determine if the followin... | False. To get the 95% confidence interval for (pfemale - pmale), all we have to do is to swap the bounds of the confidence interval for (pmale - pfemale) and take their negatives: (-0.06,-0.02). |
The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per y... | False. The confidence interval includes 0. |
The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per y... | False. We are 95% confident that 16% fewer to 2% Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year. |
The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per y... | False. As the confidence level decreases the width of the confidence level decreases as well. |
The United States federal government shutdown of 2018–2019 occurred from December 22, 2018 until January 25, 2019, a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that 48% of those who make less than $40,000 per year and 55% of those who make $40,000 or more per y... | True. |
Exercise 6.11 presents the results of a poll evaluating support for a generically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. Calculate a 95% confidence interval for the difference between the proportion of Democrats and Independe... | Standard error: SE = sqrt((0.79(1-0.79)/347) + (0.55(1-0.55)/617)) = 0.03 So 0.79 – 0.55 ± 1.96 x 0.03 = (0.181, 0.299) We are 95% confident that the proportion of Democrats who support the plan is 18.1% to 29.9% higher than the proportion of Independents who support the plan. |
Exercise 6.11 presents the results of a poll evaluating support for a generically branded “National Health Plan” in the United States. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it ... | True. |
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of ... | Before calculating the confidence interval we should check that the conditions are satisfied. 1. Independence: Both samples are random, and 11,545 < 10% of all Californians and 4,691 < 10% of all Oregonians, therefore how much one Californian sleeps is independent of how much another Californian sleeps and how much one... |
Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 ... | The hypotheses are: H0: pCA = pOR, HA: pCA ≠ pOR We have confirmed in Exercise ?? that the independence condition is satisfied but we need to recheck the success-failure condition using p^pool and expected counts. successCA = nCA x pCA = 11,545 x 0.08 = 923.6 ≈ 924; successOR = nOR x pOR = 4,691 x 0.088 = 412.8 ≈ 413; ... |
In July 2008 the US National Institutes of Health announced that it was stopping a clinical study early because of unexpected results. The study population consisted of HIV-infected women in sub-Saharan Africa who had been given single dose Nevaripine (a treatment for HIV) while giving birth, to prevent transmission of... | The hypotheses are as follows: H0: pN = pL: There is no difference in virologic failure rates between the Nevaripine and Lopinavir groups. HA: pN ≠ pL: There is some difference in virologic failure rates between the Nevaripine and Lopinavir groups. First, check conditions. 1. Independence: Random assignment was used, s... |
An apple a day keeps the doctor away. A physical education teacher at a high school wanting to increase awareness on issues of nutrition and health asked her students at the beginning of the semester whether they believed the expression “an apple a day keeps the doctor away”, and 40% of the students responded yes. Thro... | No. The samples at the beginning and at the end of the semester are not independent since the survey is conducted on the same students. |
An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 6, CL = 90% | n = 6, CL = 90%, df = 6 - 1 = 5, t* = 2.02 |
An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 21, CL = 98% | n = 21, CL = 98%, df = 21 - 1 = 20, t* = 2.53 |
An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 29, CL = 95% | n = 29, CL = 95%, df = 29 - 1 = 28, t* = 2.05 |
An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t*) for the given sample size and confidence level: n = 12, CL = 99% | n = 12, CL = 99%, df = 12 - 1 = 11, t* = 3.11 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 11, T = 1.91 | n = 11 T = 1:91 df = 11 - 1 = 10 p-value = 0.085 Do not reject H0 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 17, T = −3.45 | n = 17 T = -3:45 df = 17 - 1 = 16 p-value = 0.003 Reject H0 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 7, T = 0.83 | n = 7 T = 0:83 df = 7 - 1 = 6 p-value = 0.438 Do not reject H0 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05: n = 28, T = 2.13 | n = 28 T = 2:13 df = 28 - 1 = 27 p-value = 0.042 Reject H0 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.01: n = 26, T = 2.485 | n = 26 T = 2.485 df = 26 - 1 = 25 p-value = 0.020 Do not reject H0 |
An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.01: n = 18, T = 0.5 | n = 18 T = 0.5 df = 18 - 1 = 17 p-value = 0.623 Do not reject H0 |
A 95% confidence interval for a population mean, µ, is given as (18.985, 21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t-distribution in any calculations... | The sample mean is the mid-point of the confidence interval, i.e. the average of the upper and lower bounds: x-bar = (18.985 + 21.015) / 2 = 20. The margin of error is 21.015 - 20 = 1.015. Since n = 36, df = 35, and the critical t-score is T* = 2.03. Then, 1.015 = 2.03 x s / sqrt(36); so s = 3. |
A 90% confidence interval for a population mean is (65, 77). The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard devia... | The sample mean is the mid-point of the confidence interval, i.e. the average of the upper and lower bounds: x-bar = (65 + 77) / 2 = 71. The margin of error is 77-71 = 6. Since n = 25, df = 24, and the critical t-score is T* = 1.71. Then, 6 = 1.71 x s / sqrt(25); so s = 17.54. |
You are given the following hypotheses: H0: µ = 60; HA: µ ≠ 60; We know that the sample standard deviation is 8 and the sample size is 20. For what sample mean would the p-value be equal to 0.05? Assume that all conditions necessary for inference are satisfied. | For the single tails to each be 0.025 at n - 1 = 20 - 1 = 19 degrees of freedom, T score must equal to be either -2.09 or +2.09. Then, either: -2:09 = (x-bar – 60) / (8 / sqrt(20)); x = 56.26 or 2:09 = (x-bar – 60) / (8 / sqrt(20)); x = 63.74 |
For a given confidence level, t* is larger than z*. Explain how t* being slightly larger than z* affects the width of the confidence interval. | With a larger critical value, the confidence interval ends up being wider. This makes intuitive sense as when we have a small sample size and the population standard deviation is unknown, we should have a wider interval than if we knew the population standard deviation, or if we had a large enough sample size. |
Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Evaluate Georgianna’s claim (or that the opposite might... | We will conduct a 1-sample t-test. H0: µ = 5. HA: µ ≠ 5. We'll use α = 0:05. This is a random sample, so the observations are independent. To proceed, we assume the distribution of years of piano lessons is approximately normal. SE = 2.2= / sqrt(20) = 0.4919. The test statistic is T = (4.6 - 5)=SE = -0:81. df = 20 - 1 ... |
Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Construct a 95% confidence interval for the number of y... | Using SE = 0.4919 and t* =19 = 2.093, the confidence interval is (3.57, 5.63). We are 95% confident that the average number of years a child takes piano lessons in this city is 3.57 to 5.63 years. |
Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. Do your results from the hypothesis test and the confid... | They agree, since we did not reject the null hypothesis and the null value of 5 was in then t-interval. |
Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a S... | H0: µ = 35, HA: µ ≠ 35. |
Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a S... | 1. Independence: if we can assume that these 52 officers represent a random sample (big assumption), then independence would be satisfied, but we cannot check this. 2. Normality: We don't have a plot of the distribution that we can use to check this condition. We at least have more than 30 observations, so the distribu... |
Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32 µg/l and a S... | The test statistic and the p-value can be calculated as follows: T = (124.32 – 35) / (7.74 / sqrt(52) = 17.07; df = 52 - 1 = 51 p-value = 2 x P(T > 17.07) < 0.001. The hypothesis test yields a very small p-value, so we reject H0. Given the direction of the data, there is very convincing evidence that the police office... |
A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. How large of a sample should he collect? | 10 ≥ 1:96 x 100 / sqrt(n); n = 384.16 He should survey at least 385 customers. Note that we need to round up the calculated sample size. |
The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Raina wants to use a 9... | 25 ≥ 1.65 x 250 / sqrt(n); n = 272.25. Raina should collect a sample of at least 273 students. |
The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Luke wants to use a 99... | If Luke had the same sample size as Raina but used a higher confidence level, he would end up with wider interval. To keep the width of his confidence interval the same as Raina's Luke will need a higher sample size. |
The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. Calculate the minimum ... | 25 ≥ 2.58 x 250 / sqrt(n); n = 665.64. Luke should collect a sample of at least 666 students. |
Air quality measurements were collected in a random sample of 25 country capitals in 2013, and then again in the same cities in 2014. We would like to use these data to compare average air quality between the two years. Should we use a paired or non-paired test? Explain your reasoning. | Paired, data are recorded in the same cities at two different time points. The temperature in a city at one point is not independent of the temperature in the same city at another time point. |
Determine if the following statement is true or false. If false, explain: In a paired analysis we first take the difference of each pair of observations, and then we do inference on these differences. | True. |
Determine if the following statement is true or false. If false, explain: Two data sets of different sizes cannot be analyzed as paired data. | True. |
Determine if the following statement is true or false. If false, explain: Consider two sets of data that are paired with each other. Each observation in one data set has a natural correspondence with exactly one observation from the other data set. | True. |
Determine if the following statement is true or false. If false, explain: Consider two sets of data that are paired with each other. Each observation in one data set is subtracted from the average of the other data set’s observations. | False. We find the deference of each pair of observations, and then we do inference on these divergences. |
In the following scenario, determine if the data are paired: Compare pre- (beginning of semester) and post-test (end of semester) scores of students. | Since it's the same students at the beginning and the end of the semester, there is a pairing between the datasets, for a given student their beginning and end of semester grades are dependent. |
In the following scenario, determine if the data are paired: Assess gender-related salary gap by comparing salaries of randomly sampled men and women. | Since the subjects were sampled randomly, each observation in the men's group does not have a special correspondence with exactly one observation in the other (women's) group. |
In the following scenario, determine if the data are paired: Compare artery thicknesses at the beginning of a study and after 2 years of taking Vitamin E for the same group of patients. | Since it's the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester artery thickness are dependent. |
In the following scenario, determine if the data are paired: Assess effectiveness of a diet regimen by comparing the before and after weights of subjects. | Since it's the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester weights are dependent. |
In the following scenario, determine if the data are paired: We would like to know if Intel’s stock and Southwest Airlines’ stock have similar rates of return. To find out, we take a random sample of 50 days, and record Intel’s and Southwest’s stock on those same days. | Paired, on the same day the stock prices may be dependent on external factors that affect the price of both stocks. |
In the following scenario, determine if the data are paired: We randomly sample 50 items from Target stores and note the price for each. Then we visit Walmart and collect the price for each of those same 50 items. | Paired, the prices are for the same items. |
In the following scenario, determine if the data are paired: A school board would like to determine whether there is a difference in average SAT scores for students at one high school versus another high school in the district. To check, they take a simple random sample of 100 students from each high school. | Not paired, these are two independent random samples, individual students are not matched. |
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