Math 115 Test 2 Practice Exercises

1. A 99% confidence interval is constructed for a population mean. Compared to a 95% confidence interval using the same data, the 99% interval would be: Circle one.

  1. Narrower because we need more precision for higher confidence

  2. Wider because we need a larger range to be more confident

  3. The same width because both use the same data

  4. Cannot be determined without knowing the sample size

2. A 95% confidence interval for a population mean is (42, 58). Which of the following statements is TRUE? Circle all that apply.

  1. The sample mean is 50

  2. There is a 95% probability that the true mean is between 42 and 58

  3. We are 95% confident that the true mean is between 42 and 58

  4. 95% of all sample means fall between 42 and 58

3. A movie theater surveys 180 patrons and finds that 99 prefer watching movies in 3D. A bootstrap 95% confidence interval for the true proportion who prefer 3D is (0.48, 0.62).

Based on this confidence interval, would you reject \(H_0: p = 0.5\) in favor of \(H_A: p \neq 0.5\) at the \(\alpha = 0.05\) significance level? Circle one.

  1. Yes, reject \(H_0\) because 0.5 is not in the confidence interval

  2. Yes, reject \(H_0\) because 0.5 is in the confidence interval

  3. No, fail to reject \(H_0\) because 0.5 is not in the confidence interval

  4. No, fail to reject \(H_0\) because 0.5 is in the confidence interval

4. A psychologist wants to study stress levels among college students. She stands outside the campus coffee shop during finals week and surveys the first 50 students who walk by and agree to participate.

(a) What type of sampling method is this? Circle one.

  1. Simple random sampling
  2. Convenience sampling

(b) Can her findings be generalized to all college students at her university? Circle one.

  1. Yes, because she collected a sample of students
  2. Yes, because 50 is a reasonable sample size
  3. No, because the sample may not be representative of all students at the university
  4. No, because stress levels vary by student

5. The heights of adult women are normally distributed with mean \(\mu\) = 64 inches and standard deviation \(\sigma\) = 3 inches.

(a) What percentage of adult women are between 61 and 67 inches tall?

Answer: _________%

(b) What percentage of adult women are taller than 70 inches?

Answer: _________%

For parts (c)–(e), the heights are not a whole number of standard deviations from the mean, so the 68-95-99.7 rule is not enough. Use Jamovi (or a normal table) and give each answer as a proportion rounded to three decimal places.

(c) What proportion of adult women are shorter than 60 inches?

Answer: _________

(d) What proportion of adult women have heights from 62 to 69 inches?

Answer: _________

(e) What proportion of adult women are taller than 66.5 inches?

Answer: _________

6. A factory manager inspects 80 products and finds 6 with defects. She wants to test whether the defect rate is different from 10%.

Is the success-failure condition met for using the normal model? Circle one.

  1. Yes, because the expected numbers of successes and failures are both at least 10

  2. Yes, because the sample size of 80 is large enough

  3. No, because expected successes is less than 10

  4. No, because the observed number of defects is less than 10

7. Which of the following would make a confidence interval WIDER? Circle all that apply.

  1. Increasing the sample size

  2. Decreasing the sample size

  3. Increasing the confidence level from 95% to 99%

  4. Decreasing the confidence level from 95% to 90%

8. A political analyst tests whether a candidate has majority support. In a sample of 150 likely voters, 82 support the candidate. She conducts a two-sided test: \(H_0: p = 0.5\) vs \(H_A: p \neq 0.5\).

The Z-score is 1.14 and the area to the right of Z = 1.14 is 0.126.

(a) Calculate the two-sided p-value. (Round to three decimal places.)

Answer: _________

(b) At \(\alpha = 0.05\), what is the conclusion? Circle one.

  1. Reject \(H_0\); there is convincing evidence that the proportion is different from 0.5

  2. Fail to reject \(H_0\); there is not convincing evidence that the proportion is different from 0.5

9. A medical researcher is testing whether a new screening test can detect a disease. The null hypothesis is that the test cannot detect the disease (no better than random chance).

(a) What would a Type I error mean in this context? Circle one.

  1. Concluding the test can detect the disease when it actually can
  2. Concluding the test can detect the disease when it actually cannot
  3. Concluding the test cannot detect the disease when it actually can
  4. Concluding the test cannot detect the disease when it actually cannot

(b) What would a Type II error mean in this context? Circle one.

  1. Concluding the test can detect the disease when it actually can
  2. Concluding the test can detect the disease when it actually cannot
  3. Concluding the test cannot detect the disease when it actually can
  4. Concluding the test cannot detect the disease when it actually cannot

(c) The researcher uses a significance level of \(\alpha = 0.01\). Her study is designed so that, if the screening test really can detect the disease, the chance her study will reject \(H_0\) is 0.85. (This chance is called the power of the test.)

What is the probability that she makes a Type II error if the test really can detect the disease?

Answer: _________

(d) In this study, what is the probability of a Type I error if the screening test really cannot detect the disease?

Answer: _________

(e) Which statement describes the power of this study? Circle one.

  1. The probability of concluding the test can detect the disease when it actually can
  2. The probability of concluding the test can detect the disease when it actually cannot
  3. The probability that the null hypothesis is true
  4. The probability of failing to reject \(H_0\) when it is false

10. A researcher reports: “The average GPA of students at State University is 3.2.” This value is a: Circle one.

  1. Parameter, because it describes the entire population of State University students

  2. Statistic, because it was calculated from a sample of students

  3. Parameter if it was calculated from all students, or a statistic if from a sample

  4. Neither a parameter nor a statistic

11. Determine whether each statement about confidence intervals is TRUE or FALSE. Circle one for each.

(a) Increasing the sample size will make a confidence interval narrower. True / False

(b) A 99% confidence interval is narrower than a 95% confidence interval. True / False

(c) If a 95% CI for a proportion is (0.52, 0.68), then 0.50 is a plausible value for the population proportion. True / False

12. A restaurant chain claims that 70% of customers are satisfied with their service. A regional manager suspects the true satisfaction rate in her region is lower than 70%. She surveys a random sample of 120 customers and finds that 75 report being satisfied.

Below is a null distribution showing 1000 simulated proportions under the assumption that the true satisfaction rate is 70%.

Use this histogram and the information above to answer the following questions.

(a) What is the parameter of interest in this study? Circle one.

  1. The proportion of the 120 sampled customers who reported being satisfied

  2. The true proportion of all customers in this region who are satisfied with service

  3. The 70% satisfaction rate claimed by the restaurant chain

  4. The number of satisfied customers in the sample

(b) Which pair of hypotheses is appropriate for this research question? Circle one.

  1. \(H_0: \hat{p} = 0.70\) vs. \(H_A: \hat{p} > 0.70\)

  2. \(H_0: p = 0.70\) vs. \(H_A: p < 0.70\)

  3. \(H_0: \hat{p} = 0.70\) vs. \(H_A: \hat{p} < 0.70\)

  4. \(H_0: p = 0.70\) vs. \(H_A: p > 0.70\)

(c) Calculate the sample proportion \(\hat{p}\) of customers in the sample who reported being satisfied. (Round to two decimal places.)

Answer: _________

(d) Using the histogram, estimate the p-value for this hypothesis test. (Round to three decimal places.)

Answer: _________

(e) Which of the following correctly interprets the p-value in the context of this problem? Circle one.

  1. The probability that the true satisfaction rate is 70%

  2. The probability that the true satisfaction rate is less than 70%

  3. The probability of observing a sample proportion of \(\hat{p}\) or lower, assuming the true satisfaction rate is 70%

  4. The probability of observing a sample proportion of exactly \(\hat{p}\), assuming the true satisfaction rate is less than 70%

(f) Using a significance level of \(\alpha = 0.05\), what is the appropriate decision for this hypothesis test? Circle one.

  1. Reject \(H_0\)

  2. Fail to reject \(H_0\)

  3. Accept \(H_0\)

(g) Based on your decision in part (f), which conclusion is appropriate? Circle one.

  1. There is convincing evidence that the true satisfaction rate is less than 70%.

  2. There is not enough evidence to conclude that the true satisfaction rate is less than 70%.

  3. We have proved that the true satisfaction rate is exactly 70%.

  4. There is convincing evidence that the true satisfaction rate is greater than 70%.

(h) Suppose instead the regional manager only wanted to know whether the satisfaction rate in her region is different from 70% (higher or lower). Using the same histogram, estimate the two-sided p-value. (Round to three decimal places.)

Hint: the sample proportion is 0.075 below 0.70. Count the simulated proportions that are at least that far from 0.70 in either direction, that is, at or below 0.625 and at or above 0.775. Then divide by 1000.

Answer: _________

(i) At \(\alpha = 0.05\), what would the decision be for the two-sided test in part (h)? Circle one.

  1. Reject \(H_0\)
  2. Fail to reject \(H_0\)

13. A dentist wants to estimate the proportion of patients who floss daily. She surveys a random sample of 80 patients and finds that 52 report flossing daily.

Below is a bootstrap distribution showing 1000 bootstrap proportions based on resampling from the original sample with replacement.

Use this histogram and the information above to answer the following questions.

(a) What is the parameter of interest in this study? Circle one.

  1. The proportion of the 80 sampled patients who report flossing daily

  2. The 52 patients in the sample who floss daily

  3. The true proportion of all patients who floss daily

  4. The average number of times patients floss per week

(b) Calculate the sample proportion \(\hat{p}\) of patients in the sample who report flossing daily. (Round to two decimal places.)

Answer: _________

(c) To construct a 95% confidence interval, which percentiles of the bootstrap distribution should be used? Circle one.

  1. 5th and 95th percentiles

  2. 2.5th and 97.5th percentiles

  3. 0.5th and 99.5th percentiles

  4. 25th and 75th percentiles

(d) Using the histogram, estimate the lower bound (2.5th percentile) of the 95% confidence interval. (Round to two decimal places.)

Answer: _________

(e) Using the histogram, estimate the upper bound (97.5th percentile) of the 95% confidence interval. (Round to two decimal places.)

Answer: _________

(f) Which of the following is the correct interpretation of this 95% confidence interval? Circle one.

  1. There is a 95% probability that the true proportion of patients who floss daily is in this interval.

  2. 95% of all patients floss daily.

  3. We are 95% confident that the true proportion of all patients who floss daily is in this interval.

  4. 95% of samples would have a proportion of patients who floss daily between these two values.

(g) Based on your confidence interval, is 0.60 (60%) a plausible value for the true proportion of patients who floss daily? Circle one.

  1. Yes, because 0.60 is inside the confidence interval

  2. No, because 0.60 is outside the confidence interval

  3. Yes, because 0.60 is close to our sample proportion

  4. No, because we cannot make claims about specific values from a confidence interval

(h) What does the 95% confidence level say about the method used to build the interval? Circle one.

  1. This particular interval has a 95% probability of containing the true proportion of patients who floss daily.

  2. If we took many random samples of 80 patients and built a 95% confidence interval from each one, about 95% of those intervals would capture the true proportion of patients who floss daily.

  3. About 95% of the patients in the sample floss daily.

  4. If we repeated the survey, the new sample proportion would fall in this interval 95% of the time.

14. A fitness app company claims that 50% of users complete their daily workout goal. A product manager believes the true completion rate is higher than 50%. She analyzes data from a random sample of 200 users and finds that 120 completed their daily goal.

Use the normal model and the 68-95-99.7 rule to answer the following questions.

(a) What are the null and alternative hypotheses? Circle one.

  1. \(H_0: p = 0.50\) vs. \(H_A: p < 0.50\)

  2. \(H_0: p = 0.50\) vs. \(H_A: p > 0.50\)

  3. \(H_0: p = 0.50\) vs. \(H_A: p \neq 0.50\)

  4. \(H_0: \hat{p} = 0.50\) vs. \(H_A: \hat{p} > 0.50\)

(b) Calculate the sample proportion \(\hat{p}\). (Round to two decimal places.)

Answer: _________

(c) Calculate the standard error using the null hypothesis value \(p_0 = 0.50\). (Round to two decimal places.)

Answer: _________

(d) Calculate the Z-score. (Round to two decimal places.)

Answer: _________

(e) The Z-score tells us that the sample proportion is how many standard errors above the null hypothesis value? Circle one.

  1. About 1 standard error above
  2. About 2 standard errors above
  3. About 3 standard errors above
  4. Less than 1 standard error above

(f) Using the 68-95-99.7 rule, approximately what percentage of the normal distribution is above Z = 3? Circle one.

  1. 16%
  2. 2.5%
  3. 0.15%
  4. 5%

(g) Based on your answer to part (f), is the one-sided p-value less than \(\alpha = 0.05\)? Circle one.

  1. Yes
  2. No

(h) What is the appropriate conclusion? Circle one.

  1. Reject \(H_0\). There is convincing evidence that more than 50% of users complete their daily goal.

  2. Reject \(H_0\). There is convincing evidence that exactly 60% of users complete their daily goal.

  3. Fail to reject \(H_0\). There is not convincing evidence that more than 50% of users complete their daily goal.

  4. Fail to reject \(H_0\). There is convincing evidence that 50% of users complete their daily goal.

(i) The 68-95-99.7 rule only gives an approximate p-value. Use Jamovi to find the exact one-sided p-value for your Z-score from part (d). (Round to three decimal places.)

Answer: _________

15. A school district wants to estimate the proportion of parents who support implementing a four-day school week. They survey a random sample of 250 parents and find that 140 support the proposal.

Use the normal model to construct a 95% confidence interval for the true proportion of parents who support the four-day school week. Use \(z^* = 1.96\) for the multiplier.

(a) Calculate the sample proportion \(\hat{p}\). (Round to two decimal places.)

Answer: _________

(b) Check the success-failure condition. Are both the number of successes and the number of failures at least 10? Circle one.

  1. Yes
  2. No

(c) Calculate the standard error. Use the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). (Round to three decimal places.)

Answer: _________

(d) Calculate the margin of error using \(z^* = 1.96\). (Round to three decimal places.) Hint: margin of error \(= z^* \times SE\)

Answer: _________

(e) Calculate the lower bound of the 95% confidence interval. (Round to two decimal places.)

Answer: _________

(f) Calculate the upper bound of the 95% confidence interval. (Round to two decimal places.)

Answer: _________

(g) Which of the following is the correct interpretation of this confidence interval? Circle one.

  1. There is a 95% probability that the true proportion of parents who support the four-day week is within the interval.

  2. We are 95% confident that the true proportion of all parents who support the four-day school week is within the interval.

  3. 95% of all samples would have a proportion of parents supporting the four-day week within the interval.

  4. Exactly 95% of parents have support levels within the interval.

(h) Based on this confidence interval, is there convincing evidence that a majority (more than 50%) of parents support the four-day school week? Circle one.

  1. Yes, because the entire confidence interval is above 0.50

  2. Yes, because the sample proportion 0.56 is greater than 0.50

  3. No, because the confidence interval contains 0.50

  4. No, because we cannot draw conclusions about specific values from a confidence interval

(i) The school board decides a 90% confidence interval is good enough. Use Jamovi to find the multiplier \(z^*\) for 90% confidence. (Round to three decimal places.)

Answer: _________

(j) Using the same standard error from part (c), find the lower bound of the 90% confidence interval. (Round to three decimal places.)

Answer: _________

(k) Find the upper bound of the 90% confidence interval. (Round to three decimal places.)

Answer: _________

(l) Compared to the 95% interval, the 90% interval is: Circle one.

  1. Wider, and it still contains 0.50

  2. Narrower, and it no longer contains 0.50

  3. Narrower, and it still contains 0.50

  4. The same width, but shifted to the right

16. For each study described below, determine whether it is an experiment or an observational study. Circle one for each.

(a) A hospital randomly selects 60 patients with surgical wounds from its patient records and randomly assigns them to receive either a new antibacterial dressing or a standard dressing, then records healing time.

  1. Experiment
  2. Observational study

(b) A researcher goes to a local gym and surveys 200 members who are willing to participate about their supplement use and records whether each member reported an injury in the past year.

  1. Experiment
  2. Observational study

(c) A teacher asks for student volunteers and randomly assigns half to study with flashcards and half to study with practice problems, then compares exam scores.

  1. Experiment
  2. Observational study

(d) In study (a), which variable is the explanatory variable and which is the response variable? Circle one.

  1. Explanatory: healing time; Response: type of dressing
  2. Explanatory: type of dressing (new antibacterial or standard); Response: healing time
  3. Explanatory: the 60 patients; Response: the hospital’s patient records
  4. Explanatory: average healing time in each group; Response: difference in average healing times

(e) In study (c), the response variable is: Circle one.

  1. Study method (flashcards or practice problems)
  2. Whether the student volunteered
  3. Exam score
  4. The number of students in each group

(f) In study (b), suppose the researcher finds that members who take supplements reported more injuries than members who don’t. Which of the following is the most plausible confounding variable, meaning one that could be related to both supplement use and injuries? Circle one.

  1. How many hours per week the member trains, and how intensely
  2. The number of members the researcher surveyed
  3. The color of the gym’s equipment
  4. Whether the survey was given on paper or online

17. For each of the studies described in Problem 1, answer the following questions.

(a) Can the results be generalized to a broader population? Circle one for each.

(b) Can we conclude causation from each study? Circle one for each.

18. A health researcher surveys 240 adults about their exercise habits and how frequently they get colds. The results are shown in the contingency table below.

2+ Colds/Year <2 Colds/Year Total
Regular Exercise 28 92 120
Not Regular 52 68 120
Total 80 160 240

(a) What proportion of adults who exercise regularly had 2 or more colds per year? (Round to three decimal places.)

Answer: _________

(b) What proportion of adults who do NOT exercise regularly had 2 or more colds per year? (Round to three decimal places.)

Answer: _________

(c) Based on your answers to (a) and (b), does there appear to be an association between exercise frequency and cold frequency? Briefly explain.





19. A university randomly assigns 180 students to one of two study methods for an introductory course: 90 students use Method A (active recall) and 90 use Method B (re-reading notes). At the end of the semester, the university records whether each student passed the final exam. The results are:

The researchers believe Method A leads to a higher pass rate. Below is a null distribution showing 1000 simulated differences in proportions (\(\hat{p}_A - \hat{p}_B\)) under the assumption that the two methods are equally effective. The dashed red line marks the observed difference. Values at or beyond the observed difference are highlighted.

Use the histogram and the information above to answer the following questions.

(a) Which pair of hypotheses is appropriate for this research question? Circle one.

  1. \(H_0: p_A - p_B = 0\) vs. \(H_A: p_A - p_B > 0\)

  2. \(H_0: p_A - p_B = 0\) vs. \(H_A: p_A - p_B \neq 0\)

  3. \(H_0: \hat{p}_A - \hat{p}_B = 0\) vs. \(H_A: \hat{p}_A - \hat{p}_B > 0\)

  4. \(H_0: \hat{p}_A - \hat{p}_B = 0\) vs. \(H_A: \hat{p}_A - \hat{p}_B \neq 0\)

(b) What is the observed difference in sample proportions (\(\hat{p}_A - \hat{p}_B\))? (Round to three decimal places.)

Answer: _________

(c) Using the histogram, estimate the p-value for this hypothesis test. (Round to three decimal places.)

Answer: _________

(d) Using a significance level of \(\alpha = 0.05\), what is the appropriate decision? Circle one.

  1. Reject \(H_0\)
  2. Fail to reject \(H_0\)
  3. Accept \(H_0\)

(e) Which conclusion is appropriate? Circle one.

  1. There is convincing evidence that Method A causes a higher pass rate than Method B.

  2. There is not enough evidence to conclude that Method A leads to a higher pass rate than Method B.

  3. There is convincing evidence of an association between study method and pass rate, but we cannot conclude causation.

  4. We have proved that the two methods are equally effective.

(f) Identify the variables in this study. Circle one.

  1. Explanatory: whether the student passed the final exam; Response: study method
  2. Explanatory: study method (A or B); Response: whether the student passed the final exam
  3. Explanatory: the pass rate of Method A; Response: the pass rate of Method B
  4. Explanatory: the 180 students; Response: the difference in pass rates

20. Determine whether each statement is TRUE or FALSE. Circle one for each.

(a) Permutation tests simulate the null hypothesis by shuffling the response values and recalculating the statistic many times.

True / False

(b) A bootstrap confidence interval requires knowing the population parameter in order to construct it.

True / False

(c) If a 95% bootstrap confidence interval for \(p_1 - p_2\) contains 0, we would fail to reject \(H_0: p_1 - p_2 = 0\) at the \(\alpha = 0.05\) significance level.

True / False

21. A researcher uses a bootstrap method to construct a 95% confidence interval for the difference in two population proportions (\(p_1 - p_2\)) and obtains the interval (0.02, 0.14). Which of the following is the best interpretation? Circle one.

  1. There is a 95% probability that \(p_1 - p_2\) is between 0.02 and 0.14.

  2. We are 95% confident that the true difference in population proportions (\(p_1 - p_2\)) is between 0.02 and 0.14.

  3. If we repeated this study, 95% of the time the sample difference in proportions would be between 0.02 and 0.14.

  4. Because the interval does not contain 0, we are 95% confident that \(p_1 = p_2\).

22. A medical clinic wants to know whether the type of appointment reminder affects whether patients show up. The clinic randomly assigns 240 patients: 120 receive an email reminder and 120 receive a text reminder. The results are shown below.

Showed Up Did Not Show Total
Email Reminder 82 38 120
Text Reminder 70 50 120
Total 152 88 240

Conduct a two-sided hypothesis test at \(\alpha = 0.05\) to determine whether there is a difference in show-up rates between the two reminder types. Let \(\hat{p}_E\) be the proportion who showed up in the email group and \(\hat{p}_T\) be the proportion who showed up in the text group.

(a) Which pair of hypotheses is appropriate? Circle one.

  1. \(H_0: p_E - p_T = 0\) vs. \(H_A: p_E - p_T \neq 0\)

  2. \(H_0: p_E - p_T = 0\) vs. \(H_A: p_E - p_T > 0\)

  3. \(H_0: \hat{p}_E - \hat{p}_T = 0\) vs. \(H_A: \hat{p}_E - \hat{p}_T \neq 0\)

  4. \(H_0: \hat{p}_E - \hat{p}_T = 0\) vs. \(H_A: \hat{p}_E - \hat{p}_T > 0\)

(b) Calculate the pooled proportion. (Round to three decimal places.)

Answer: _________

(c) What is the observed difference in sample proportions (\(\hat{p}_E - \hat{p}_T\))? (Round to three decimal places.)

Answer: _________

(d) Calculate the standard error. (Round to four decimal places.)

Answer: _________

(e) Calculate the Z-statistic. (Round to two decimal places.)

Answer: _________

(f) Use Jamovi to find the p-value. (Round to three decimal places.)

Answer: _________

(g) At \(\alpha = 0.05\), what is the appropriate decision? Circle one.

  1. Reject \(H_0\); there is convincing evidence of a difference in show-up rates.

  2. Fail to reject \(H_0\); there is not convincing evidence of a difference in show-up rates.

  3. Accept \(H_0\); there is convincing evidence that the show-up rates are the same.

(h) Below is the histogram of 1000 bootstrapped differences of proportions

Find the bounds of a 95% confidence interval based on this histogram.

The lower bound of the 95% confidence interval.

Answer: _________

The upper bound of the 95% confidence interval.

Answer: _________

23. A city wants to compare composting rates between urban and suburban households. Random samples of 200 urban and 200 suburban households are surveyed. The results are shown below.

Compost Do Not Compost Total
Urban 96 104 200
Suburban 68 132 200
Total 164 236 400

Construct a 95% confidence interval for the difference in population proportions (\(p_U - p_S\)).

(a) What is the observed difference in sample proportions (\(\hat{p}_U - \hat{p}_S\))? (Round to three decimal places.)

Answer: _________

(b) Calculate the standard error. (Round to four decimal places.)

Answer: _________

(c) What is \(z^*\) for a 95% confidence interval? Use Jamovi. (Round to three decimal places.)

Answer: _________

(d) Calculate the lower bound of the 95% confidence interval. (Round to three decimal places.)

Answer: _________

(e) Calculate the upper bound of the 95% confidence interval. (Round to three decimal places.)

Answer: _________

Answer Key

1

Answer: B

A 99% confidence interval must be wider than a 95% interval because we need a larger margin of error to achieve higher confidence.

2

Correct answers: A and C

  • A: The midpoint of the interval (42, 58) is the sample mean (42 + 58) / 2 = 50

  • C: Correct CI interpretation: we are 95% confident the population mean lies in the interval.

3

Bootstrap 95% CI: (0.48, 0.62)

Answer: D

Because 0.50 is inside the interval, we fail to reject H₀ at α = 0.05.

4

  1. Convenience sampling

  2. No, the sample may not represent all students.

Students surveyed near a coffee shop during finals week may have unusually high stress.

5

  1. 68%

  2. 2.5%

Explanation:

  • 61–67 = μ ± 1σ → about 68%
  • 70 = μ + 2σ → upper tail ≈ 2.5%

  1. z = (60 − 64)/3 = −1.33 → P(X < 60) ≈ 0.091

  2. z = (62 − 64)/3 = −0.67 and z = (69 − 64)/3 = 1.67 → P(62 < X < 69) ≈ 0.952 − 0.252 = 0.700

  3. z = (66.5 − 64)/3 = 0.83 → P(X > 66.5) ≈ 0.202

In Jamovi, enter mean = 64 and SD = 3 in the normal distribution tool. You can also standardize first and use a standard normal (mean 0, SD 1).

6

Expected successes under the null:

np₀ = 80(0.10) = 8

Answer: C

The success-failure condition requires ≥10 expected successes and failures.

7

Correct answers: B and C

  • Decreasing sample size increases variability.
  • Increasing confidence level increases the critical value.

Both make intervals wider.

8

Two-sided p-value:

p = 2(0.126) = 0.252

  1. 0.252

  2. Fail to reject H₀

Because p > 0.05.

9

  1. Type I error

Concluding the test detects the disease when it actually cannot.

  1. Type II error

Concluding the test cannot detect the disease when it actually can.

  1. P(Type II error) = 1 − power = 1 − 0.85 = 0.15

  2. P(Type I error) = α = 0.01

  3. i: power is the probability of correctly rejecting H₀ when it is false. Here that means concluding the test can detect the disease when it really can. Option (iv) is the Type II error probability, which is 1 − power.

10

Correct answer: C

It is a parameter if computed from the entire population, or a statistic if from a sample.

11

  1. True

  2. False

  3. False

Explanation:

  • Increasing sample size narrows intervals.
  • Higher confidence makes intervals wider.
  • Since 0.50 is outside (0.52, 0.68), it is not plausible.

12

  1. Parameter:

True proportion of customers satisfied in the region

  1. Hypotheses

H₀: p = 0.70

Hₐ: p < 0.70

p̂ = 75 / 120 = 0.625

Estimated p-value ≈ 0.037

Correct interpretation: iii

Since p < 0.05, reject H₀.

There is convincing evidence the satisfaction rate is below 70%.

Two-sided p-value: count the simulated proportions ≤ 0.625 and ≥ 0.775, then divide by 1000.

Two-sided p-value ≈ 0.086. This is roughly double the one-sided p-value, because the null distribution is roughly symmetric.

Fail to reject H₀. The two-sided p-value is above 0.05, even though the one-sided test rejected H₀, because a two-sided test counts both tails.

13

  1. Parameter:

True proportion of patients who floss daily

p̂ = 52 / 80 = 0.65

Use 2.5% and 97.5% percentiles

Lower bound ≈ 0.55

Upper bound ≈ 0.75

Correct interpretation:

We are 95% confident the population proportion lies in the interval.

Yes.

0.60 is inside the interval, so it is plausible.

ii: the confidence level describes the long-run success rate of the method. If we built 95% intervals from many random samples, about 95% of them would capture the true proportion. Option (i) is wrong because any one interval either contains p or it doesn’t. Option (iv) describes sample proportions, not intervals.

14

H₀: p = 0.50 Hₐ: p > 0.50

p̂ = 120 / 200 = 0.60

SE = sqrt(0.5×0.5 / 200) = 0.035

z = (0.60 − 0.50) / 0.035 ≈ 2.83

≈ 3 standard errors above

Area above z = 3 ≈ 0.15%

Yes, p-value < 0.05.

Reject H₀.

Evidence completion rate exceeds 50%.

Area to the right of z = 2.83 under the standard normal curve ≈ 0.002. This agrees with the 68-95-99.7 estimate: it is well below 0.05.

15

p̂ = 140 / 250 = 0.56

Successes = 140 Failures = 110

Condition satisfied.

SE = sqrt(0.56×0.44 / 250) = 0.031

ME = 1.96 × 0.031 = 0.061

Lower bound

0.56 − 0.061 = 0.499

Upper bound

0.56 + 0.061 = 0.621

CI ≈ (0.50, 0.62)

Correct interpretation:

We are 95% confident the population proportion lies in the interval.

No.

Since 0.50 is inside the interval, a majority is not established.

z* = 1.645. This leaves 5% in each tail, so it is the 95th percentile of the standard normal.

ME = 1.645 × 0.0314 ≈ 0.052

Lower bound = 0.56 − 0.052 ≈ 0.508 (0.509 if you used SE rounded to 0.031)

Upper bound = 0.56 + 0.052 ≈ 0.612 (0.611 if you used SE rounded to 0.031)

ii: the 90% interval (0.508, 0.612) is narrower and lies entirely above 0.50. Lowering the confidence level makes the interval narrower, and here that changes the conclusion about a majority.

16

  1. Experiment

  2. Observational study

  3. Experiment

Experiments assign treatments; observational studies only observe existing behavior.

  1. ii: Explanatory: type of dressing; Response: healing time

  2. iii: Exam score

The explanatory variable is the one the researcher assigns or uses to explain differences. The response variable is the outcome that is measured.

  1. i: Training time and intensity. People who train harder are more likely to take supplements and more likely to get injured. That could create an association between supplements and injuries even if supplements cause no injuries.

17

Generalization

Study A: Yes Study B: No Study C: No

Causation

Study A: Yes Study B: No Study C: Yes

Random sampling supports generalization. Random assignment supports causation.

18

28 / 120 = 0.233

52 / 120 = 0.433

Yes, there appears to be an association.

Adults who do not exercise regularly have a higher cold rate.

19

H₀: pA − pB = 0 Hₐ: pA − pB > 0

54/90 − 42/90 = 0.133

Estimated p-value ≈ 0.03

Reject H₀.

Evidence Method A increases pass rate.

ii: Explanatory: study method (A or B); Response: whether the student passed the final exam

20

  1. True

  2. False

  3. True

21

Correct answer: B

We are 95% confident the difference in population proportions lies between 0.02 and 0.14.

22

H₀: pE − pT = 0 Hₐ: pE − pT ≠ 0

p_pool = 152 / 240 = 0.633

0.683 − 0.583 = 0.100

SE ≈ 0.062

z ≈ 1.61

p ≈ 0.107

Fail to reject H₀.

There is not convincing evidence of a difference.

Lower bound: about 0.025

Upper Bound: about 0.225

23

0.48 − 0.34 = 0.14

SE ≈ 0.047

z* = 1.96

Lower bound ≈ 0.048

Upper bound ≈ 0.232