Title: Inference for One Proportion
Course: Question Banks

Q: A 2016 Gallup survey received 1,671 responses from randomly selected college and university faculty members in the U.S.

Faculty diverge around their perceptions of online courses based on whether they have experience teaching online; 32% of those with online teaching experience and only 13% without it agree that online and in-person courses can achieve equivalent outcomes at any institution. </p

An objective of this study is to ________ .
- estimate and compare proportions between two groups (We want to estimate the proportions of college and university faculty members in the U.S. that believe in the effectiveness of online learning. The variable is categorical (“support” or “oppose”), so we summarize with a proportion.)
- test a claim about a difference in proportions between two groups (We want to estimate the percentage of college and university faculty members in the U.S. with a given opinion. If the question contained a claim, it might say “do more than 60% have this opinion” or something like that.)
- test a claim about a difference in means between two groups (We want to estimate the percentage of college and university faculty members in the U.S. with a given opinion, so we summarize with a proportion rather than a mean. Furthermore, if the question contained a claim, it might say “do more than 60% have this opinion” or something like that.)
- estimate and compare means between two groups (The variable is categorical (“support” or “oppose”), so we summarize with a proportion rather than a mean.)
A: estimate and compare means between two groups

Q: The BBC repeatedly polled its readers regarding support of Brexit, a proposed UK withdrawal from the EU which was voted in a 23 June 2016 referendum. (Source)

The question of interest was to ________
- estimate a population proportion (We want to estimate the percentage of U.K. adults with a given opinion. The variable is categorical (“favor” or “oppose”), so we summarize with a proportion but a stronger conclusion is of interest.)
- estimate a population mean (The variable is categorical (“favor” or “oppose”), so we summarize with a proportion rather than a mean.)
- test a claim about a population mean (We want to estimate the percentage of U.K. adults with a given opinion. If the question contained a claim, it might say “do more than 60% have this opinion” or something like that. Also, the variable is categorical (“favor” or “oppose”), so we summarize with a proportion rather than a mean.)
- test a claim about a population proportion (We want to estimate the percentage of U.K. adults with a given opinion. The variable is categorical (“favor” or “oppose”), and the question of interest is whether the referendum would pass (have a proportion greater than 50%).)
A: test a claim about a population proportion

Q: In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount.

Calculate the 99% confidence interval for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount for U.S. economic problems.
- (0.504, 0.556) (This is the 90% confidence interval.)
- (0.499, 0.561) (This is the 95% confidence interval.)
- The confidence interval cannot be determined from the information given. (In order to determine the confidence interval we only need the sample proportion and the sample size. Both are given in this problem.)
- (0.489, 0.571) (The estimated standard error is 0.0158 so the confidence interval is given by: 0.53 ± 2.576 (0.0158). Note that your answer may vary slightly due to rounding.)
A: (0.489, 0.571)

Q: In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount.

Using these data, we calculated a 99% confidence interval of (0.489, 0.571) for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount.

Does the confidence interval indicate that a majority of all American adults blame Barack Obama a great deal or a moderate amount for U.S. economic problems?
- Yes, because the entire interval is above 0.5. (The 99% confidence interval is (0.489, 0.571). The interval extends both below and above 0.5.)
- Yes, because 53% is a majority. (A majority of the sample has this opinion. But this does not mean a majority of the population has this opinion.)
- No, because a sample of 1004 adults can’t tell us about the entire population. (Even though 1004 seems like a small number compared to the entire population, as long as the sample is chosen randomly, it should be representative of the population. This will allow us to use the results of the sample to make statements about the population.)
- No, because the interval extends below 0.5, meaning that it is reasonably possible that a minority of all American adults feel this way. (Good Job! In order to be convinced that a majority of American adults blame Barack Obama a great deal or a moderate amount, the entire confidence interval must be above 0.5. The 99% confidence interval we calculated in the previous problem extends down to 0.489, meaning that a minority is a reasonable possibility.)
A: No, because the interval extends below 0.5, meaning that it is reasonably possible that a minority of all American adults feel this way.

Q: Imagine that you read a newspaper article that reports on a survey finding that 72% of Americans support outlawing the common airline practice of overbooking seats (overbooking means selling more seats than are actually available, assuming that some people won’t show up for the flight). The researchers note that the margin of error is ± 5% using 95% confidence.

What does this margin of error tell you about the results of the survey?
- We are 95% confident that the results are off by 5%. (The problem with this statement is the phrase “the results are off by 5%.” The margin of error tells us the maximum distance of the population proportion from the observed sample proportion with a given level of confidence.)
- 5% of the respondents gave a fake answer. (The margin of error only gives us information about the expected variability in sample proportions with a given level of confidence. It does not give us information about other sources of error such as fake answers.)
- We are certain that the proportion of all Americans who support outlawing overbooking is between 67% and 77%. (The margin of error is not a definite maximum distance, but the most likely maximum distance of the population proportion from the observed sample proportion.)
- We can be 95% confident that the proportion of all Americans who support outlawing overbooking is within 5% of 72%. (We can be confident that the population proportion is no further from the sample proportion than the margin of error.)
A: We can be 95% confident that the proportion of all Americans who support outlawing overbooking is within 5% of 72%.

Q: In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount.

Using these data, we calculated a 99% confidence interval of (0.489, 0.571) for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount.

What does this 99% confidence level tell us?
- There is a 99% chance that this particular interval contains the population proportion. (We cannot make a probability statement about a particular interval because there is no random event. We already found the interval. We can only make probability statements about the long run behavior of random samples or confidence intervals from random samples.)
- The interval will contain 99% of all sample proportions. (This statement is only true if the sample proportion equals the population proportion. The confidence level is a probability statement about the long run behavior of random samples or confidence intervals from random samples. It is not a statement about a percentage of sample proportions.)
- Of confidence intervals with this margin of error, 99% will contain the sample proportion. (The confidence interval is a sample proportion ± a margin of error. So every confidence interval contains a sample proportion at its center.)
- Of confidence intervals with this margin of error, 99% will contain the population proportion. (Since 99% of sample proportions will fall within 2.576 standard errors of p, 99% of the time the confidence interval will contain p.)
A: Of confidence intervals with this margin of error, 99% will contain the population proportion.

Q: We survey a random sample of American River College students and ask if they drink coffee on a regular basis. The 90% confidence interval for the proportion of all American River College students who drink coffee on a regular basis is (0.262, 0.438). What will be true about the 95% confidence interval for these data?
- The 95% confidence interval is narrower than the 90% confidence interval. (The value of Z_c is larger for the 95% confidence interval (Z_c = 1.96 vs. Z_c = 1.645). So the margin of error is larger for the 95% confidence interval. How does this affect the width of the confidence interval?)
- The two intervals will have the same width. (The confidence level determines the value of Z_c. So the confidence level impacts the margin of error and thus impacts the width of the interval.)
- It is impossible to say which interval will be wider. (The confidence level determines the value of Z_c. So the confidence level impacts the margin of error and thus impacts the width of the interval.)
- The 95% confidence interval is wider than the 90% confidence interval. (The value of Z_c is larger for the 95% confidence interval (Z_c = 1.96 vs. Z_c = 1.645). So the 95% confidence interval is wider.)
A: The 95% confidence interval is wider than the 90% confidence interval.

Q: Imagine that you are a researcher conducting a survey for a client. You are trying to decide what sample size you need. What would the main advantage be of a larger sample size?
- A larger sample size allows more people to give their input. (You are right that a larger sample size does allow more people to be included in the sample, but that is not the primary reason to increase the sample size.)
- A larger sample size means the sampling method isn’t as important. (The sampling method is critically important regardless of the sample size.)
- A larger sample size gives a higher confidence level. (Any confidence level can be used with any sample size, as long as there are at least 10 successes and failures.)
- A larger sample size gives a narrower confidence interval. (The reason for using a larger sample size is to decrease the margin of error.)
A: A larger sample size gives a narrower confidence interval.

Q: At Valencia College, are male students more likely to be smokers than female students?

What is the null hypothesis in this case?
- The proportion of male Valencia College students who are smokers is greater than the proportion of female Valencia College students who are smokers. (Remember that the null hypothesis is a statement of equality between the two parameters. This statement is the alternative hypothesis.)
- The proportion of male Valencia College students who are smokers is less than the proportion of female Valencia College students who are smokers. (Remember that the null hypothesis is a statement of equality between the two parameters. The statement in this answer is an inequality.)
- There is a difference between the proportion of male Valencia College students who are smokers and the proportion of female Valencia College students who are smokers. (Remember that the null hypothesis is a statement of equality between the two parameters. The statement in this answer is an inequality.)
- There is no difference between the proportion of male Valencia College students who are smokers and the proportion of female Valencia College students who are smokers. (The null hypothesis in this case would be that the proportion of male students who smoke and the proportion of female students who smoke are the same.)
A: There is no difference between the proportion of male Valencia College students who are smokers and the proportion of female Valencia College students who are smokers.

Q: At Valencia College, are male students more likely to be smokers than female students?

What is the alternative hypothesis in this case?
- There is no difference between the proportion of male Valencia College students who are smokers and the proportion of female Valencia College students who are smokers. (The alternative hypothesis is an inequality statement. Saying there is no difference means the proportion of male students who smoke is equal to the proportion of female students who smoke. This is the null hypothesis.)
- The proportion of male Valencia College students who are smokers is less than the proportion of female Valencia College students who are smokers. (Although this is an inequality, we are interested in whether male students are more likely to be smokers than female students.)
- There is a difference between the proportion of male Valencia College students who are smokers and the proportion of female Valencia College students who are smokers. (Although this is an inequality, we are interested in whether male students are more likely to be smokers than female students. This statement only says there is a difference between the two.)
- The proportion of male Valencia College students who are smokers is greater than the proportion of female Valencia College students who are smokers. (If male students are more likely to be smokers than female students, then the proportion of males who smoke is greater than the proportion of females who smoke.)
A: The proportion of male Valencia College students who are smokers is greater than the proportion of female Valencia College students who are smokers.

Q: Suppose that you’re in charge of a Coca-Cola manufacturing plant that produces 16 oz bottles of Coke. The bottles are filled by a machine, and you want to make sure that the machine is filling the bottles with an average of 16 oz of Coke. In order to do this, you periodically select a random sample of bottles to check.

Which of the following is an appropriate statement of the null hypothesis?
- The machine is not filling the bottles with the proper amount of Coke. The average is not 16 ounces. (We assume the machines are filling correctly and then see if the sample provides evidence otherwise.)
- The machine is not putting enough Coke in the bottles. The average is less than 16 ounces. (We assume the machines are filling correctly and then see if the sample provides evidence otherwise.)
- The machine fills the bottles with the proper amount of Coke. The average is 16 ounces. (We assume the machines are filling correctly and then see if the sample provides evidence otherwise. So H_0 states that the average is 16 ounces.)
A: The machine fills the bottles with the proper amount of Coke. The average is 16 ounces.

Q: A commonly accepted metric is that college students study 2 hours outside of class for every hour in class. This means a full-time student taking 15 units (hours of class) studies for 30 hours per week. An educator suspects this figure is different now than in the past.

- H_0: The average time full-time college students study outside of class per week is 30 hours.
- H_a: The average time full-time college students study outside of class per week is not 30 hours.

To substantiate her claim, the educator randomly selects 1,500 college students and finds that they study an average of 27 hours per week with a standard deviation of 1.7 hours.

In order to assess the evidence, we need to determine:
- How likely it is to observe a mean number of hours of studying outside of class per week that is different from 30 hours per week. (Your choice asks how likely it is that the alternative hypothesis is true. Recall that in hypothesis testing, in order to assess the evidence, we need to find how likely is it to get data like those observed assuming that the null hypothesis is true.)
- How likely it is in a random sample of 1,500 students to observe that the average number of hours spent per week studying outside of class is at most 27 hours, or at least 33 hours. (Recall that in hypothesis testing, in order to assess the evidence, we need to find how likely is it to get data like those observed assuming that the null hypothesis is true.)
- How likely it is in a random sample of 1,500 students to observe that the mean amount of hours studying outside of class per week is 27 hours or less. (You have only addressed the possibility that different means "less than." Different could also mean "greater than." Also, we need to find how likely it is to get data like those observed assuming that the null hypothesis is true.)
- How likely it is that the mean number of hours per week spent by students studying outside of class is 30. (Your choice asks how likely it is that the null hypothesis is true. Recall that in hypothesis testing, in order to assess the evidence, we need to find how likely is it to get data like those observed assuming that the null hypothesis is true.)
- How likely it is in a random sample of 1,500 students to observe students studying an average of at most 27 or at least 33 hours per week outside of class, if the mean number is actually 30 hours per week. (Good job! Indeed, in hypothesis testing, in order to assess the evidence, we need to find how likely it is to get data like those observed assuming that the null hypothesis is true.)
A: How likely it is in a random sample of 1,500 students to observe students studying an average of at most 27 or at least 33 hours per week outside of class, if the mean number is actually 30 hours per week.

Q: The following two hypotheses are tested:

- H_0: The average number of miles driven per year is 12,000.
- H_a: The average number of miles driven per year is less than 12,000.

In a survey, 1,600 randomly selected drivers were asked the number of miles they drive yearly. Based on the results, the P-value = 0.068.

Comment: Throughout this activity, use a 0.05 (5%) significance level.

The fact that the P-value is .068 means that:
- There is a probability of .068 of observing data like those observed. (The P-value is a conditional probability. We must state that we are assuming that the null is true when we describe the P-value.)
- There is a probability of .068 that the average number of miles driven per year is 12,000. (This statement says in essence, “the P-value is the probability that the null hypothesis is true.” We cannot make a probability statement about the value of the parameter because the parameter is fixed. We can only make probability statements about random events.)
- There is a .932 probability that the average number of miles driven per year is less than 12,000. (This statement says in essence, “the P-value is the probability that the alternative hypothesis is true.” We cannot make a probability statement about the value of the parameter because the parameter is fixed. We can only make probability statements about random events, like sample means.)
- There is a probability of .068 (reasonably likely) of observing data like those observed if the average number of miles driven per year is 12,000. (Good job! Indeed, the P-value is the probability of observing data like those observed assuming that the null hypothesis, H_o, is true.)
A: There is a probability of .068 (reasonably likely) of observing data like those observed if the average number of miles driven per year is 12,000.

Q: The following two hypotheses are tested:

- H_0: The average number of miles driven per year is 12,000.
- H_a: The average number of miles driven per year is less than 12,000.

In a survey, 1,600 randomly selected drivers were asked the number of miles they drive yearly. Based on the results, the P-value = 0.068.

Comment: Throughout this activity, use a 0.05 (5%) significance level.

Based upon the p-value, which of the following conclusions can be made?
- The data provide significant evidence that the average number of miles driven per year is less than 12,000. (Note that the p-value is more than 0.05.)
- The data provide significant evidence that the average number of miles driven per year is 12,000. (Recall that we never "accept" the null hypothesis; however, we can "not reject" it instead.)
- The data provide significant evidence that the alternative hypothesis is more likely than the null hypothesis. (It should be noted that the statement that you chose is actually statistically inaccurate. Note that the number of miles driven per year is a parameter (µ), and therefore a fixed, unknown quantity about which we are making inferences. We can talk about the probability that a random variable takes certain value(s), but not that a fixed quantity does.)
- The data do not provide significant evidence that the average number of miles driven per year is less than 12,000. (Since the p-value is > 0.05, we cannot accept the alternative hypothesis. We could have also stated that we do not reject the null hypothesis (that the data does not provide significant evidence to reject that the average number of miles driven per year is 12,000).)
A: The data do not provide significant evidence that the average number of miles driven per year is less than 12,000.

Q: The following two hypotheses are tested:

- H_0: The average number of miles driven per year is 12,000.
- H_a: The average number of miles driven per year is less than 12,000.

In a survey, 1,600 randomly selected drivers were asked the number of miles they drive yearly. Based on the results, the P-value = 0.068.

Comment: Throughout this activity, use a 0.05 (5%) significance level.

Suppose that the p-value was not given, but instead the following conclusion was made: data provides significant evidence that the average number of miles driven per year is less than 12,000. Which of the following could have been the p-value?
- −1.96 (Since a p-value is a probability, it must be between 0 and 1.)
- 0.05 (This p-value wouldn't provide significant evidence to accept the alternative hypothesis. The p-value must be less than 0.05 for this.)
- 0.068 (Look at the first question, which did have this p-value. What was the conclusion?)
- 0.032 (The p-value must be less than 0.05 to provide enough evidence to reject the null hypothesis and accept the alternative.)
A: 0.032

Q: The following two hypotheses are tested:

- H_0: The average number of miles driven per year is 12,000.
- H_a: The average number of miles driven per year is less than 12,000.

In a survey, 1,600 randomly selected drivers were asked the number of miles they drive yearly. Based on the results, the P-value = 0.068.

Comment: Throughout this activity, use a 0.05 (5%) significance level.

When would you conclude that the average number of miles driven per year is less than 12,000?
- Never (While we can never accept the null hypothesis, we can accept the alternative hypothesis if there is sufficient evidence.)
- When the p-value is large (more than 0.05) (If the p-value is > 0.05 we would not reject the null hypothesis. This question is asking when we would reject the null hypothesis.)
- When the average number of miles is less than 12,000 (The key issue is: is the average number of miles a significant amount less than 12,000. Sometimes, even though the average is less than 12,000, we would not reject the null hypothesis in favor of the alternative hypothesis.)
- When the p-value is small (less than 0.05) (We can accept H_a if there is sufficient evidence, namely that the p-value < 0.05.)
A: When the p-value is small (less than 0.05)

Q: The null hypothesis claims that the proportion of community college students who work full-time is 0.45. The alternative hypothesis is that this proportion is greater than 0.45.

Which gives the most convincing evidence against this null hypothesis and in support of the alternative?
- A random sample of 1,000 community college students in which 300 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. The sampling distribution is centered at 0.45. Sample proportions that support the alternative are greater than 0.45. Here the sample proportion is 0.3. This does not support the alternative hypothesis since it is less than 0.45.)
- A random sample of 1,000 community college students in which 500 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. The sampling distribution is centered at 0.45. Sample proportions that support the alternative are greater than 0.45. Sample proportions further from 0.45 give stronger evidence against the null and in support of the alternative. Here the sample proportion is 0.5. This supports the alternative hypothesis but is not the most convincing evidence.)
- A random sample of 1,000 community college students in which 700 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. The sampling distribution is centered at 0.45. Here the sample proportion is 0.7. This is the largest sample proportion above 0.45, so it gives the most convincing evidence against the claim that p = 0.45 and in support of the claim p > 0.45.)
A: A random sample of 1,000 community college students in which 700 report working full-time.

Q: The null hypothesis claims that the proportion of community college students who work full-time is 0.45. The alternative hypothesis is that this proportion is greater than 0.45.

Which sample proportion will have the smallest P-value?
- A random sample of 1,000 community college students in which 500 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. A sample proportion further from 0.45 in the direction of the alternative hypothesis will be more rare. So it has the smaller P-value. Here the sample proportion is 0.5. Did you compute the other sample proportions?)
- A random sample of 1,000 community college students in which 600 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. A sample proportion further from 0.45 in the direction of the alternative hypothesis will be more rare. So it has the smaller P-value. Here the sample proportion is 0.6. Did you compute the other sample proportions?)
- A random sample of 1,000 community college students in which 700 report working full-time. (Since the samples are all the same size, we can visualize the sampling distribution. A sample proportion further from 0.45 in the direction of the alternative hypothesis will be more rare. So it has the smaller P-value. So the sample proportion 0.7 has the smallest P-value.)
A: A random sample of 1,000 community college students in which 700 report working full-time.

Q: In a hypothesis test, we choose a 5% level of significance. Which P-value is statistically significant at this level?
- 0.08 (This P-value is larger than the significance level of 0.05.)
- 0.06 (This P-value is larger than the significance level of 0.05.)
- 0.04 (This P-value is smaller than the significance level of 0.05.)
A: 0.04

Q: Toys intended for use by infants are regularly inspected to see if they contain illegal concentrations of lead (lead can cause brain damage in infants that consume it). This can be thought of as a hypothesis test with the following hypotheses.

H_0: The toy is safe.

H_a: The toy is not safe.

Is the following statement a Type I or Type II error?

The sample suggests that the toy is safe, but it actually is not safe.
- Type I (The statement describes a situation where we fail to reject a false null hypothesis. A Type I error requires that we reject the null hypothesis.)
- Type II (The statement describes a situation where we fail to reject a false null hypothesis. This is a Type II error.)
A: Type II

Q: Toys intended for use by infants are regularly inspected to see if they contain illegal concentrations of lead (lead can cause brain damage in infants that consume it). This can be thought of as a hypothesis test with the following hypotheses.

H_0: The toy is safe.

H_a: The toy is not safe.

Is the following statement a Type I or Type II error?

The sample suggests that the toy is not safe, but it actually is safe.
- Type II (The inspector’s conclusion here is that H_0 is false (the toy is not safe). But H_0 is actually true (the toy is safe). A Type II error requires that we fail to reject H_0.)
- Type I (The inspector’s conclusion here is that H_0 is false (the toy is not safe). But H_0 is actually true (the toy is safe).)
A: Type I

Q: A lifeguard finds an odd-looking mole on his arm and goes to the dermatologist to get it checked out. The doctor removes the mole and sends it to the lab to be checked for melanoma (a dangerous form of skin cancer). The results of this test could be considered a hypothesis test with the following hypotheses:

H_0: The lifeguard does not have melanoma.

H_a: The lifeguard does have melanoma.

Which of the following statements is considered a Type I error?
- The lifeguard does have melanoma, but the test result shows he does not have melanoma. (The statement describes a situation in which the null hypothesis is false, but we fail to reject it. This is not a Type I error.)
- The lifeguard does have melanoma, and the test result shows he does have melanoma. (The statement indicates that there is no error.)
- The lifeguard does not have melanoma, and the test result shows he does not have melanoma. (The statement indicates that there is no error.)
- The lifeguard does not have melanoma, but the test result shows he does have melanoma. (The statement describes a situation in which the null hypothesis is true, but we reject it. This is a Type I error.)
A: The lifeguard does not have melanoma, but the test result shows he does have melanoma.

Q: A lifeguard finds an odd-looking mole on his arm and goes to the dermatologist to get it checked out. The doctor removes the mole and sends it to the lab to be checked for melanoma (a dangerous form of skin cancer). The results of this test could be considered a hypothesis test with the following hypotheses:

H_0: The lifeguard does not have melanoma.

H_a: The lifeguard does have melanoma.

Which of the following statements is considered a Type II error?
- The lifeguard does not have melanoma, but the test result shows he does have melanoma. (The statement describes a situation in which the null hypothesis is true, but we reject it. This is not a Type II error..)
- The lifeguard does not have melanoma, and the test result shows he does not have melanoma. (The statement indicates that there is no error.)
- The lifeguard does have melanoma, and the test result shows he does have melanoma. (The statement indicates that there is no error.)
- The lifeguard does have melanoma, but the test result shows he does not have melanoma. (The statement describes a situation in which the null hypothesis is false, but we fail to reject it. This is a Type II error.)
A: The lifeguard does have melanoma, but the test result shows he does not have melanoma.

Q: For the following scenario, identify the variable as either quantitative or categorical.

A poll of students at your college asks each student to give an estimate of the number of military causalities that have occurred since the United States invaded Iraq in March of 2003.
- categorical (The variable is the estimate of the number of military casualties. It is quantitative. So this type of scenario will be analyzed using means, not proportions.)
- quantitative (The variable is the estimate of the number of military casualties. It is quantitative. So this type of scenario will be analyzed using means, not proportions.)
A: quantitative

Q: For the following scenario, identify the variable as either quantitative or categorical.

A poll of students at your college shows that 2 out of 3 do not support continuing military intervention in Iraq.
- quantitative (The variable is support (or not) for continuing military intervention. It is categorical. So this type of scenario will be analyzed using proportions.)
- categorical (The variable is support (or not) for continuing military intervention. It is categorical. So this type of scenario will be analyzed using proportions.)
A: categorical

Q: For the following scenario, identify the variable as either quantitative or categorical.

A local newspaper claims that 67% of the county's residents support a bond measure. We conduct a phone survey and find that 85 out of the 150 people contacted support the measure.
- quantitative (The variable is support (or not) for a bond measure. It is categorical. So this type of scenario will be analyzed using proportions.)
- categorical (The variable is support (or not) for a bond measure. It is categorical. So this type of scenario will be analyzed using proportions.)
A: categorical

Q: For each scenario, identify the variable as either quantitative or categorical.

A local newspaper claims that the county's residents commute an average of 18 miles each way to work. We conduct a phone survey in which we ask the number of miles the respondent drives each way to work.
- categorical (The variable is number of miles commuted each way to work. It is quantitative. So this type of scenario will be analyzed using means, not proportions.)
- quantitative (The variable is number of miles commuted each way to work. It is quantitative. So this type of scenario will be analyzed using means, not proportions.)
A: quantitative

Q: According to the American Association of Community Colleges, 23% of community college students receive federal grants. The California Community College Chancellor’s Office anticipates that the percentage is smaller for California community college students. They collect a sample of 1,000 community college students in California and find that 210 received federal grants.

For which group are we making a hypothesis?
- the population of community college students (We do not need to make a hypothesis about community college students. For this population, we already know that 23% receive federal grants.)
- a sample of community college students in California (We do not need to make a hypothesis about a sample. We can always calculate the proportion using the sample data.)
- the population of community college students in California (We are trying to answer a question about California community college students.)
A: the population of community college students in California

Q: According to the American Association of Community Colleges, 23% of community college students receive federal grants. The California Community College Chancellor’s Office anticipates that the percentage is smaller for California community college students. They collect a sample of 1,000 community college students in California and find that 210 received federal grants.

We want to find out if p for California is 0.23 or if it is smaller than the national number.

Which are the correct hypotheses for this scenario?
- H_0: p = 0.23, H_a: p < 0.210 (The null and alternative hypotheses always use the same value.)
- H_0: p = 0.210, H_a: p > 0.210 (The hypotheses are statements about a population proportion. 0.210 is a sample proportion.)
- H_0: p = 0.23, H_a: p ≠ 0.23 (The null hypothesis is correct, but the alternative is not. Reread the scenario. What word(s) help you choose among >, <, or ≠?)
- H_0: p = 0.23, H_a: p < 0.23 (The alternative hypothesis reflects the claim that the Chancellor’s Office anticipates that “the percentage is smaller.”)
A: H_0: p = 0.23, H_a: p < 0.23

Q: According to the American Association of Community Colleges, 23% of community college students receive federal grants. The California Community College Chancellor’s Office anticipates that the percentage is smaller for California community college students. They collect a sample of 1,000 community college students in California and find that 210 received federal grants.

We want to find out if p for California is 0.23 or if it is smaller than the national number: H_0: p = 0.23, H_a: p < 0.23

What does p represent in the hypotheses?
- the proportion of community college students who received federal grants (We do not need to make a hypothesis about the proportion of community college students who receive federal grants. We already know that this is 23%.)
- the proportion of 1,000 community college students who received federal grants (We do not need to make a hypothesis about the proportion of the 1,000 California community college students who received federal grants. We can calculate it: 210 / 1,000 = 0.210.)
- the number of federal grant recipients attending community colleges in California (The hypothesis is about a proportion, not a number or count.)
- the proportion of California community college students who received federal grants (The population of interest is California community college students. The variable is “receive federal grants” (yes/no).)
A: the proportion of California community college students who received federal grants

Q: In 2007, a Gallup poll estimated that 45% of U.S. adults rated their financial situation as “good.” We want to know if the proportion is smaller this year. We gather a random sample of 100 U.S. adults this year and find that 39 rate their financial situation as “good.”

After carrying out a hypothesis test for p = 0.45 compared to p < 0.45, we obtain a p-value of 0.114. Is the following statement about the p-value valid or invalid?

There is an 11.4% chance that the percentage of adults who rate their financial situation as “good” is still 45%.
- valid (This statement really says “the p-value is the probability that the null hypothesis is true.” Probability describes random events. The null hypothesis is either true or not true. The p-value is a probability about the results from the random sample.)
- invalid (This statement really says “the p-value is the probability that the null hypothesis is true.” Probability describes random events. The null hypothesis is either true or not true. The p-value is a probability about the results from the random sample.)
A: invalid

Q: In 2007, a Gallup poll estimated that 45% of U.S. adults rated their financial situation as “good.” We want to know if the proportion is smaller this year. We gather a random sample of 100 U.S. adults this year and find that 39 rate their financial situation as “good.”

After carrying out a hypothesis test for p = 0.45 compared to p < 0.45, we obtain a p-value of 0.114. Is the following statement about the p-value valid or invalid?

There is an 11.4% chance that the percentage of adults who rate their financial situation as “good” is now less than 45%.
- valid (This statement really says “the p-value is the probability that the alternative hypothesis is true.” Probability describes random events. The alternative hypothesis is either true or not true. The p-value is a probability about the results from the random sample.)
- invalid (This statement really says “the p-value is the probability that the alternative hypothesis is true.” Probability describes random events. The alternative hypothesis is either true or not true. The p-value is a probability about the results from the random sample.)
A: invalid

Q: In 2007, a Gallup poll estimated that 45% of U.S. adults rated their financial situation as “good.” We want to know if the proportion is smaller this year. We gather a random sample of 100 U.S. adults this year and find that 39 rate their financial situation as “good.”

After carrying out a hypothesis test for p = 0.45 compared to p < 0.45, we obtain a p-value of 0.114. Is the following statement about the p-value valid or invalid?

If the percentage of adults who rate their financial situation as “good” is still 45%, there is an 11.4% chance that a random sample of size 100 would have a sample percentage of 39% or lower.
- invalid (The null hypothesis is “the percentage of adults who rate their financial situation as “good” is still 45%. If this is true, there is only a 11.4% chance that we will see a sample percentage as low as 39% in a sample of size 100.)
- valid (This is a good description of the p-value. The null hypothesis is “the percentage of adults who rate their financial situation as “good” is still 45%” If this is true, there is only a 11.4% chance that we will see a sample percentage as low as 39% in a sample of size 100.)
A: valid

Q: In 2009 news organizations reported that 47% of U.S. adults favored providing a legal way for illegal immigrants already in the U.S. to become citizens. In 2010, the percentage with this opinion rose to 50%. Suppose that a current poll of 100 randomly selected U.S. adults shows this percentage is higher than 50% this year.

A researcher conducts a hypothesis test with the data from the current poll. She finds that the increase this year is not statistically significant. What conclusion can we draw?
- The researcher must have misinterpreted the results. There should be a significant increase. (Higher does not mean “significantly” higher in the statistical meaning of that word. “Not statistically significant” means it is not unusual to see random samples of this size with percentages that vary above 50% by the amount seen in the current poll.)
- It must be true that 50% of the public still has this opinion this year. (We can never say that the null hypothesis is true. We can only say that we do not have enough strong enough data to reject it.)
- The sample size may be too small to detect a statistically significant increase. (This year a majority of the whole population may actually favor providing a legal way for illegal immigrants to become citizens. But the increase is not recognized as significant by the hypothesis test because the sample size is too small. Larger samples will vary less, so the hypothesis test will be more sensitive to detecting an increase.)
A: The sample size may be too small to detect a statistically significant increase.

Q: The following is an excerpt from cancerguide.org, a website started by Steve Dunn when he discovered he had advanced kidney cancer at the age of 32. His goal for the website is to help cancer patients understand the technical medical information. Much of his website focuses on making sense of the statistical information reported with medical trials.

“It’s also important to realize that ‘statistically significant difference’ does not mean ‘big difference.’ If two treatments are very similar in outcome, but not exactly the same, you can find a statistically significant difference by just testing enough people. In general, the more people you include in a trial, the smaller a difference is needed before that difference proves to be statistically significant. So if in some trial, treatment A has a cure rate of 52% and treatment B 54%, then even if they tested enough patients to make this difference ‘statistically significant,’ you are not likely to decide that B is really much better, and very likely other characteristics of A and B such as side effects would guide your choice between them.”

Suppose a common drug used to treat high blood pressure causes nausea in 30% of patients who take it. A clinical trial with a new drug for high blood pressure shows a “statistically significant” decrease in the percentage of patients who experience nausea. What can we conclude?
- A much smaller percentage (much lower than 30%) of patients have side effects with the new drug. (We know the percentage with side effects is lower than 30%, but we do not know how much lower. Statistically significant difference does not mean “big difference.”)
- The new drug is much less dangerous than the old drug. (This conclusion is vague and inaccurate. Statistically significant difference does not mean “big difference.” Also, a word like “dangerous” is not a good choice for describing a side effect like nausea.)
- A smaller percentage of patients have side effects with the new drug, but we do not know how much smaller. (Statistically significant difference does not mean “big difference.”)
A: A smaller percentage of patients have side effects with the new drug, but we do not know how much smaller.

Q: Imagine that you are a researcher working on a cure for Alzheimer’s disease. You develop a drug that you think could help people suffering from Alzheimer’s. To assess the effectiveness of the drug, you conduct a controlled study with a sample of 1200 patients. You find that 10% of patients taking a placebo report improvements, but your new drug shows improvement in an even higher proportion of patients, and the difference is statistically significant.

Which of the following best describes how you should interpret these results?
- Because of the large size of the sample, these results are strong evidence that the new drug will help a much larger proportion of population. (Large samples will have less variability. Therefore, the statistically significant results could come from a small improvement in the percentage who find improvement from the new drug.)
- With a large sample, statistically significant results suggest a large improvement over the placebo. (Large samples will have less variability. Therefore, the statistically significant results could come from a small improvement in the percentage who are helped by the new drug.)
- Regardless of the sample size, a statistically significant result means there is a meaningful difference in Alzheimer’s improvement.. (The sample size can impact whether a statistically significant result corresponds to a difference that matters. For large samples, a small increase may be statistically significant, but that difference may not be meaningful. For small samples, a larger increase would be needed to produce statistically significant results.)
- With a large sample, statistically significant results may actually be only a small improvement over the placebo. (Large samples will have less variability. Therefore, the statistically significant results could come from a small improvement in the percentage who are helped by the new drug.)
A: With a large sample, statistically significant results may actually be only a small improvement over the placebo.