Title: Linking Probability to Statistical Inference
Course: Question Banks

Q: What is the sample proportion of females for the sample shown?
- 0.60 (This is the population proportion. This sample has a different proportion of females.)
- 17 (You have correctly identified the number of females, but you need to find the proportion of females out of 25.)
- 0.47 (The sample proportion is number of females divided by number in the sample. You may have found the ratio of males to females by dividing 8 by 17.)
- 0.68 (The sample proportion is 17/25 = 0.68.)
A: 0.68

Q: We collect random samples of 25 students at a time and calculate the proportion of females in each sample. The standard deviation of p-hats is approximately 0.10. Which of the following is a plausible standard deviation for samples of 100?
- 0.40 (You chose a larger standard deviation; but in the simulation, the sample proportions for n = 100 were more tightly grouped about the population proportion.)
- 0.10 (You chose the same standard deviation; but in the simulation, the sample proportions for n = 100 were more tightly grouped about the population proportion.)
- 0.05 (If the sample size is increased, the standard deviation will decrease because larger samples have less variability.)
A: 0.05

Q: The SAT-Verbal scores of a sample of 300 students at a particular university had a mean of 592 and standard deviation of 73.

According to the university's reports, the SAT-Verbal scores of all its students had a mean of 580 and a standard deviation of 110.

Which of the following is a statistic?
- 300 (300 is the sample size, which neither a parameter nor a statistic.)
- 580 (580 is the population mean, which is a parameter.)
- 110 (110 is the population standard deviation, which is a parameter.)
- 592 (Indeed, 592 is the sample mean, which is a statistic.)
A: 592

Q: The SAT-Verbal scores of a sample of 300 students at a particular university had a mean of 592 and standard deviation of 73.

According to the university's reports, the SAT-Verbal scores of all its students had a mean of 580 and a standard deviation of 110.

Which of the following is a parameter?
- 300 (300 is the sample size, which is neither a parameter nor a statistic.)
- 592 (592 is the sample mean, which is a statistic.)
- 73 (73 is the sample standard deviation, which is a statistic.)
- 110 (Indeed, 110 is the population standard deviation, which is a parameter.)
A: 110

Q: The SAT-Verbal scores of a sample of 300 students at a particular university had a mean of 592 and standard deviation of 73.

According to the university's reports, the SAT-Verbal scores of all its students had a mean of 580 and a standard deviation of 110.

x-bar, μ, s, and σ are:
- 73, 110, 592, 580 (In particular, 73 is not the sample mean. It is the sample standard deviation (the other three are wrong too). It seems like you are confusing mean and standard deviation.)
- 580, 592, 110, 73 (It seems like you are confusing the sample quantities with the population quantities.)
- 300, 592, 73, 110 (In particular, 300 is the sample size, and 592 is the sample mean.)
- 592, 580, 73, 110 (For the sample of 300 students, the mean score is 592 (x-bar) and the standard deviation of these scores is 73 (s). For the population of all students at the university, the mean score is 580 (µ) and the standard deviation of scores in the population is 110 (σ).)
A: 592, 580, 73, 110

Q: During the 2008 presidential election, newspapers reported that Obama received 40% of the white male vote. If this is true, what is the probability that a random sample of 1,000 white male voters had less than 35% voting for Obama?
- About 5% (Were you thinking that the error is 5% since 40% – 35% = 5%? You need to find the standard error and the Z-score like we did in the previous example.)
- About 8%
- We cannot determine the probability because a normal model is not a good fit for the sampling distribution. (The normality conditions were met in the previous problem for samples of size 200. So larger samples (in this case 1,000) will also meet these conditions.)
- About 0% (The standard error is about 0.015. The Z-score is about -3.33. P(Z < -3.33) is about 0.)
A: About 0%

Q: The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population.

Suppose that we purchase a small bag of Skittles. Assume this size bag always has 40 candies. In this particular bag 10 are green. What is the probability that a randomly selected bag of this size has 10 or more green candies? (Note: the answer may vary slightly due to rounding in the calculation of the Z-score.)
- about 0.05 or 5% (You may be confusing the error (p-hat minus p) with the probability. Follow the steps in the example. Start by checking normality conditions.)
- about 0.21 or 21% (You correctly calculated the probability. But you forgot to check the normality conditions. A normal model is not a good fit here. The expected number of successes is not 10 (20% of 40 is only 8). So we cannot trust probability calculations from a normal model.)
- about 0.79 or 79% (You may have found the probability that there are 10 or fewer green candies. Note that we wanted the probability of 10 or MORE green candies. More importantly, you forgot to check the normality conditions. A normal model is not a good fit here. The expected number of successes is not 10 (20% of 40 is only 8). So we cannot trust probability calculations from a normal model.)
- We cannot determine the probability because a normal model is not a good fit for the sampling distribution. (A normal model is not a good fit here. The number of expected successes is only 8 (20% of 40 is only 8).)
A: We cannot determine the probability because a normal model is not a good fit for the sampling distribution.

Q: The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population.

Suppose that we purchase a mid-size bag of Skittles. Assume this size bag always has 200 candies. In this particular bag, 50 are green. What is the probability that a randomly selected bag of this size has 50 or more green candies? (Note: the answer may vary slightly due to rounding in the calculation of the Z-score.)
- Between about 1.77 and 1.79% (These are Z-scores, not probabilities. You need to find the probability (the area under the standard normal curve).)
- About 96.2 or 96.3% (This answer suggests that you calculated the Z-score correctly. But we want the probability that a bag has 50 or MORE green candies. So we want the area to the right of Z = 1.79. You gave the area to the left.)
- We cannot determine the probability because a normal model is not a good fit for the sampling distribution. (The conditions for use of a normal model are met. Expected successes and failures are at least 10. We expect 20% green in 200; np = 200(0.20) = 40. We expect 80% not green in 200; n(1 - p) = 200(0.80) = 160.)
- About 3.7 or 3.8% (The standard error is about 0.028. The Z-score is about 1.79. The area to the right of Z = 1.79 is about 3.8%.)
A: About 3.7 or 3.8%

Q: In the 2008 presidential election newspapers reported that Obama received 40% of the white male vote. If this is true, what is the probability that a random sample of 1,000 white male voters estimates this percentage within ±3% error?
- about 2 to 3% (Your answer suggests that you correctly calculated the SE and the Z-score. (SE is about 0.015 and Z-score is about 2.) You may have found the probability in one tail. This is the probability that error is MORE THAN 3%. We want the probability that error is WITHIN ± 3%. This means the error is less than 3% in either direction.)
- about 4 to 6% (Your answer suggests that you correctly calculated the SE and the Z-score. (SE is about 0.015 and Z-score is about 2.) Did you find the probability in the two tails? For this problem, this is incorrect. We want the probability that error is WITHIN ± 3%. This means the error is less than 3% in either direction.)
- We cannot determine the probability because a normal model is not a good fit for the sampling distribution. (Normality conditions are met. The expected number of both successes and failures is greater than 10. We expect 40% of 1,000 to vote for Obama; np = 1,000(0.40) = 400. We expect 60% of 1,000 not to vote for Obama; n(1 – p) = 1,000(0.60) = 600.)
- about 95 to 96% (The SE is about 0.015 and Z-score is about 2. Probability of more than 3% error is 0.0228. So the probability that error is WITHIN ± 3% is the middle area. 1 – 2 x 0.0228 is 0.9544, which is between 95 and 96%.)
A: about 95 to 96%

Q: The results of a Gallup Poll were presented in the article “Obesity Quadruples Diabetes Risk for Most US Adults” on April 6, 2017. The article reports “By their mid-to-late 30s, 9.3% of adults who are obese have been diagnosed with diabetes.” Suppose the standard error (at the 95% confidence level) is approximately ±0.001.

Which interval is the approximate 95% confidence interval?
- 0.092 to 0.094 (You calculated 0.093 ± 0.001. The margin of error in a 95% confidence interval is about 2 standard errors. The 95% confidence interval is (sample proportion) ± 2(standard errors).)
- 0.093 to 0.094 (You calculated 0.093 + 0.001. The 95% confidence interval is (sample proportion) ± 2(standard errors).)
- 0.093 to 0.095 (You calculated 0.093 + 2(0.001). You need to calculate (sample proportion) ± 2(standard errors).)
- 0.091 to 0.095 (The margin of error in a 95% confidence interval is about 2 standard errors. The 95% confidence interval is 0.093 ± 2(0.001).)
A: 0.091 to 0.095

Q: In a 2017 Gallup article “In US, Water Pollution Worries Highest Since 2001”, 63% of respondents indicated they “worried a great deal about pollution of drinking water”. The Gallup survey methods section notes that “a random sample of 1,018 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia …. [and] the margin of sampling error is ±4 percentage points at the 95% confidence level”.

Which of the following statements is correct about the 95% confidence interval?
- We are 95% confident that between 55% and 71% of all adults nationwide are “worried a great deal about pollution of drinking water”. (You calculated 63% ± 2(4%), confusing the confidence interval of (sample statistic) ± (margin of error) with one that incorporates the standard error instead of the margin of error: (sample statistic) ± 2(standard error).)
- We are 95% confident that 63% of the 1,018 adults in the sample are “worried a great deal about pollution of drinking water”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error).)
- We are 95% confident that 63% of adults nationwide are “worried a great deal about pollution of drinking water”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error).)
- We are 95% confident that between 59% and 67% of all adults nationwide are “worried a great deal about pollution of drinking water”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error), which is 63% ± 4%.)
A: We are 95% confident that between 59% and 67% of all adults nationwide are “worried a great deal about pollution of drinking water”.

Q: In the March 16, 2017 Gallup article “Nearly Two in Three in US Say Winter Was Warmer This Year”, 64% of respondents reported winter temperatures in their local area were "warmer than usual”.

The following is an excerpt from the Poll Methodology and Definitions section of the article.

“”Results for this Gallup poll are based on telephone interviews conducted March 1-5, 2017, with a random sample of 1,018 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia. For results based on the total sample of national adults, the margin of sampling error is ±4 percentage points at the 95% confidence level.

Which of the following statements is correct about the 95% confidence interval?
- We are 95% confident that between 56% and 72% of all adults nationwide believe winter temperatures in their local area were "warmer than usual”. (You calculated 63% ± 2(4%), confusing the confidence interval of (sample statistic) ± (margin of error) with one that incorporates the standard error instead of the margin of error: (sample statistic) ± 2(standard error).)
- We are 95% confident that 63% of the 1,018 adults in the sample believe winter temperatures in their local area were "warmer than usual”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error).)
- We are 95% confident that 63% of adults nationwide believe winter temperatures in their local area were "warmer than usual”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error).)
- We are 95% confident that between 60% and 68% of all adults nationwide believe winter temperatures in their local area were "warmer than usual”. (The phrase “95% confident” refers to the confidence interval of (sample statistic) ± (margin of error), which is 63% ± 4%.)
A: We are 95% confident that between 60% and 68% of all adults nationwide believe winter temperatures in their local area were "warmer than usual”.

Q: With data from the 2009 National Health Interview Survey, the Centers for Disease Control and Prevention estimated that 9.4% of U.S. children had asthma. Is the percentage higher for children living in poverty?

Suppose we select a random sample of 50 poor children and find that 9 of the 50 have asthma.

The conditions are not met for use of a normal model, so we ran a simulation with [latex]p=0.094[/latex].

Do the data suggest that the percentage of poor children with asthma is greater than 9.4%?
- Yes, because 18% of the sample of poor children have asthma and this almost twice 9.4%. (This statement is true, but the relative size of the sample proportion is not what is important. We have to determine if the sample result is likely or unlikely to occur if 9.4% of all poor children have asthma.)
- No, because only 3.6% of the poor children have asthma. (3.6% does not represent the children with asthma. This is the percentage of samples with proportions of 0.18 or higher.)
- No, because a sample proportion of 0.18 or higher is not unusual if the proportion of all poor children with asthma is 0.094. (A sample proportion of 0.18 or higher is unusual if p=0.094. This only occurs about 3.6% of the time in random samples of 50 children. This is so unusual that we say the data gives strong evidence that the proportion of all poor children with asthma is greater than 0.094.)
- Yes, because 18% in a random sample of 50 is unusual if 9.4% of all poor children have asthma. (If 9.4% of poor children have asthma, we will rarely see 18% or more with asthma in a random sample of 50. So we conclude that the data is strong enough to support our claim that the asthma rate for poor children is greater than 9.4%.)
A: Yes, because 18% in a random sample of 50 is unusual if 9.4% of all poor children have asthma.