Statistics Exam 3 Study Guide: Step-by-Step Guidance

Q1. Given a 98% confidence interval for the percentage of U.S. adults who bought a new car in the last year is (0.06, 0.094):

• a. Compute $\hat{p}$ (the sample proportion).

• b. Compute the margin of error.

Background

Topic: Confidence Intervals for Proportions

This question tests your understanding of how to interpret a confidence interval for a population proportion and how to extract the sample proportion and margin of error from the interval endpoints.

Key Terms and Formulas

• $\hat{p}$: Sample proportion (point estimate for the population proportion).

• Margin of Error (E): Half the width of the confidence interval.

Confidence interval for a proportion: $\left( \hat{p} - E, \hat{p} + E \right)$

Step-by-Step Guidance

1. Identify the lower and upper bounds of the confidence interval: $0.06$ and $0.094$.

2. Recall that the confidence interval is centered at the sample proportion $\hat{p}$, so $\hat{p}$ is the midpoint of the interval.

3. Calculate $\hat{p}$ by averaging the lower and upper bounds: $\hat{p} = \frac{0.06 + 0.094}{2}$.

4. To find the margin of error, subtract the lower bound from the upper bound and divide by 2: $E = \frac{0.094 - 0.06}{2}$.

Try solving on your own before revealing the answer!

Final Answers:

$\hat{p} = 0.077$

Margin of error $E = 0.017$

We found the midpoint for $\hat{p}$ and half the interval width for the margin of error.

Q2. In a survey of 902 randomly selected U.S. adults, 702 said they voted in the most recent presidential election.

• a. Determine a 95% confidence interval for the proportion of people who say they voted in the last election. (No need to interpret.)

• b. Actual voting records show that 61% of adults did, in fact, vote in the last election. What does this imply about the 702 people who said they voted?

Background

Topic: Confidence Intervals for Proportions and Interpretation

This question tests your ability to construct a confidence interval for a population proportion and to compare survey results to actual data.

Key Terms and Formulas

• Sample proportion: $\hat{p} = \frac{x}{n}$

• Standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• Confidence interval: $\hat{p} \pm z^* \cdot SE$

• For 95% confidence, $z^* \approx 1.96$

Step-by-Step Guidance

1. Calculate the sample proportion: $\hat{p} = \frac{702}{902}$.

2. Compute the standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{902}}$.

3. Multiply the standard error by the critical value for 95% confidence ($z^* = 1.96$): $E = 1.96 \times SE$.

4. Construct the confidence interval: $\hat{p} \pm E$.

Try solving on your own before revealing the answer!

Final Answer:

Sample proportion $\hat{p} \approx 0.778$

95% CI: $(0.751, 0.805)$

This means the proportion of people who say they voted is higher than the actual 61%, suggesting some people may overreport voting.

Q3. The data below are the hours 20 randomly selected OCC students spent studying for a math test: 0, 3, 0, 2, 5, 9, 1, 3, 2, 0, 2, 3, 4, 6, 1, 2, 0, 1, 2, 5

• What requirements are needed to construct a confidence interval for the mean hours spent studying? Does this data satisfy those? Explain.

• Regardless, construct a 95% confidence interval for the mean hours spent studying and interpret it.

Background

Topic: Confidence Intervals for the Mean (Small Sample)

This question tests your understanding of the assumptions required for constructing a confidence interval for the mean and your ability to compute it using sample data.

Key Terms and Formulas

• Requirements: Random sample, normality (or large enough sample for Central Limit Theorem), unknown population standard deviation.

• Sample mean: $\bar{x}$

• Sample standard deviation: $s$

• Confidence interval: $\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}$

• $t^*$: Critical value from t-distribution with $n-1$ degrees of freedom.

Step-by-Step Guidance

1. Check requirements: Is the sample random? Is the sample size large enough (n ≥ 30) or is the data approximately normal?

2. Calculate the sample mean $\bar{x}$ and sample standard deviation $s$ from the data.

3. Find the appropriate $t^*$ value for 95% confidence and $n-1$ degrees of freedom ($n=20$).

4. Compute the standard error: $SE = \frac{s}{\sqrt{n}}$.

5. Construct the confidence interval: $\bar{x} \pm t^* \cdot SE$.

Try solving on your own before revealing the answer!

Final Answer:

Requirements: The sample is random, but the data is skewed with several zeros, so normality may be questionable. However, with n=20, the t-interval is often used if there are no extreme outliers.

95% CI: (2.01, 3.69) hours

This means we are 95% confident that the true mean hours spent studying is between 2.01 and 3.69 hours.

Q4. A hospital samples 30 patients' blood sugar levels: mean = 6.1 mmol/L, s = 0.7 mmol/L. Construct a 95% confidence interval for the standard deviation of body temperatures. What assumption must we make? Is it reasonable? Explain.

Background

Topic: Confidence Interval for a Standard Deviation (Chi-Square Distribution)

This question tests your understanding of constructing a confidence interval for a population standard deviation and the assumptions required for the chi-square method.

Key Terms and Formulas

• Assumption: The data must be a random sample from a normally distributed population.

• Confidence interval for variance: $\left( \frac{(n-1)s^2}{\chi^2_{\text{upper}}}, \frac{(n-1)s^2}{\chi^2_{\text{lower}}} \right)$

• For standard deviation, take the square root of the interval endpoints.

• $n = 30$, $s = 0.7$

Step-by-Step Guidance

1. State the assumption: The sample must come from a normally distributed population.

2. Find the degrees of freedom: $df = n - 1 = 29$.

3. Look up the critical chi-square values for 95% confidence and 29 degrees of freedom: $\chi^2_{\text{lower}}$ and $\chi^2_{\text{upper}}$.

4. Plug values into the formula for the confidence interval for variance.

5. Take the square root of the interval endpoints to get the confidence interval for the standard deviation.

Try solving on your own before revealing the answer!

Final Answer:

Assumption: The sample is random and the population is normally distributed. This is often reasonable for biological measurements, but should be checked if possible.

95% CI for standard deviation: (0.58, 0.89) mmol/L

We used the chi-square distribution and the sample variance to construct the interval.

Q5. An article claims 21% of Michigan high school students drop out. In a sample of 665 students, 129 dropped out. Using the p-value method, is there sufficient evidence to suggest the true proportion is less than 21%?

Background

Topic: Hypothesis Testing for a Proportion (One-Sample z-Test)

This question tests your ability to set up and conduct a hypothesis test for a population proportion using the p-value method.

Key Terms and Formulas

• Null hypothesis: $H_0: p = 0.21$

• Alternative hypothesis: $H_1: p < 0.21$

• Significance level: $\alpha$ (to be specified)

• Test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

• p-value: Probability of observing a test statistic as extreme or more extreme than the one calculated, under $H_0$.

Step-by-Step Guidance

1. State the hypotheses: $H_0: p = 0.21$, $H_1: p < 0.21$.

2. Calculate the sample proportion: $\hat{p} = \frac{129}{665}$.

3. Compute the standard error: $SE = \sqrt{\frac{0.21 \times 0.79}{665}}$.

4. Calculate the z-test statistic: $z = \frac{\hat{p} - 0.21}{SE}$.

5. Find the p-value corresponding to your z-statistic (for a left-tailed test).

Try solving on your own before revealing the answer!

Final Answer:

Test statistic $z \approx -2.13$, p-value $\approx 0.0166$

Since p-value < 0.05, we reject the null hypothesis and conclude there is evidence the true dropout rate is less than 21%.

The p-value is the probability of observing a sample proportion as low as or lower than 129/665, assuming the true proportion is 0.21.

Q6. You want to study the percentage of people who contract a side effect from drug X. Using a significance level of 0.01 and a margin of error no more than 5%, how large should your sample be?

Background

Topic: Sample Size Determination for Proportions

This question tests your ability to calculate the required sample size to achieve a desired margin of error for a confidence interval for a proportion.

Key Terms and Formulas

• Margin of error: $E$

• Critical value for 99% confidence: $z^* \approx 2.576$

• Sample size formula: $n = \frac{z^{*2} p^* (1-p^*)}{E^2}$

• If $p^*$ is unknown, use $p^* = 0.5$ for maximum sample size.

Step-by-Step Guidance

1. Identify the desired margin of error $E = 0.05$ and confidence level (99%, so $z^* = 2.576$).

2. Since $p^*$ is unknown, use $p^* = 0.5$ for a conservative estimate.

3. Plug values into the sample size formula: $n = \frac{(2.576)^2 \times 0.5 \times 0.5}{(0.05)^2}$.

Try solving on your own before revealing the answer!

Final Answer:

Required sample size $n \approx 664$

This ensures the margin of error is no more than 5% with 99% confidence.

Q7. A newspaper claims adults tell a lie 8 times a day. A study records the following number of lies per day: 9, 11, 13, 22, 9, 8, 8, 8, 18, 8, 17. What assumption must be made for a hypothesis test to be valid? Is it reasonable? Justify.

• Regardless, use the p-value method to test if there is sufficient evidence to conclude the newspaper is wrong.

Background

Topic: Hypothesis Testing for a Mean (Small Sample, t-Test)

This question tests your understanding of the assumptions for a t-test and your ability to perform a hypothesis test for a mean.

Key Terms and Formulas

• Assumption: The sample is random and the population is normally distributed.

• Null hypothesis: $H_0: \mu = 8$

• Alternative hypothesis: $H_1: \mu \neq 8$

• Test statistic: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$

• Degrees of freedom: $n-1$

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 8$, $H_1: \mu \neq 8$.

2. Check the assumption: Is the sample random? Is the data approximately normal (no extreme outliers)?

3. Calculate the sample mean $\bar{x}$ and sample standard deviation $s$.

4. Compute the test statistic: $t = \frac{\bar{x} - 8}{s/\sqrt{n}}$.

5. Find the p-value for your calculated t-statistic with $n-1$ degrees of freedom.

Try solving on your own before revealing the answer!

Final Answer:

Assumption: The sample is random and the data is approximately normal. With some high values, normality may be questionable, but t-test is robust to moderate departures.

Test statistic $t \approx 2.36$, p-value $\approx 0.041$

Since p-value < 0.05, there is evidence the mean number of lies differs from 8.

Q8. The standard deviation for the height of women is 2.6 in. A random sample of 10 supermodels had s = 1.1 in. Test the claim that the standard deviation of the height of supermodels is less than the whole population of women using $\alpha = 0.10$.

Background

Topic: Hypothesis Test for a Standard Deviation (Chi-Square Test)

This question tests your ability to perform a hypothesis test for a population standard deviation using the chi-square distribution.

Key Terms and Formulas

• Null hypothesis: $H_0: \sigma = 2.6$

• Alternative hypothesis: $H_1: \sigma < 2.6$

• Test statistic: $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$

• Degrees of freedom: $n-1$

• p-value: Area to the left of the test statistic in the chi-square distribution

Step-by-Step Guidance

1. State the hypotheses: $H_0: \sigma = 2.6$, $H_1: \sigma < 2.6$.

2. Calculate the test statistic: $\chi^2 = \frac{(10-1) \times (1.1)^2}{(2.6)^2}$.

3. Find the p-value: Use the chi-square distribution with 9 degrees of freedom to find the probability to the left of your test statistic.

4. Compare the p-value to $\alpha = 0.10$ to decide whether to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

Test statistic $\chi^2 \approx 1.62$, p-value $\approx 0.12$

Since p-value > 0.10, we fail to reject the null hypothesis. There is not enough evidence to conclude the standard deviation is less.

Q9. A random sample of 100 people in Michigan found 18 say winter is their favorite season. A similar sample of 85 people in Tennessee found 28 say winter is their favorite. Find a 95% confidence interval for the difference in proportions and interpret it.

Background

Topic: Confidence Interval for the Difference of Proportions

This question tests your ability to construct and interpret a confidence interval for the difference between two population proportions.

Key Terms and Formulas

• Sample proportions: $\hat{p}_1 = \frac{18}{100}$, $\hat{p}_2 = \frac{28}{85}$

• Standard error: $SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

• Confidence interval: $(\hat{p}_1 - \hat{p}_2) \pm z^* \cdot SE$

• For 95% confidence, $z^* = 1.96$

Step-by-Step Guidance

1. Calculate the sample proportions for each group.

2. Compute the standard error for the difference in proportions.

3. Multiply the standard error by the critical value $z^* = 1.96$ to get the margin of error.

4. Construct the confidence interval: $(\hat{p}_1 - \hat{p}_2) \pm$ margin of error.

Try solving on your own before revealing the answer!

Final Answer:

95% CI: (-0.28, -0.04)

This means the proportion of people who prefer winter is significantly lower in Michigan than in Tennessee.

Q10. Researchers studied the effect of color on cognitive tasks. Word recollection results:

• Red: n = 35, mean = 15.89, s = 5.9

• Blue: n = 36, mean = 12.31, s = 5.48

Use a 0.05 significance level to test the claim that the samples are from populations with the same mean.

Background

Topic: Two-Sample t-Test for Means (Independent Samples)

This question tests your ability to perform a hypothesis test comparing the means of two independent samples.

Key Terms and Formulas

• Null hypothesis: $H_0: \mu_1 = \mu_2$

• Alternative hypothesis: $H_1: \mu_1 \neq \mu_2$

• Test statistic: $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

• Degrees of freedom: Use the formula for unequal variances (Welch's t-test)

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu_1 = \mu_2$, $H_1: \mu_1 \neq \mu_2$.

2. Calculate the difference in sample means: $15.89 - 12.31$.

3. Compute the standard error: $SE = \sqrt{\frac{5.9^2}{35} + \frac{5.48^2}{36}}$.

4. Calculate the t-statistic: $t = \frac{15.89 - 12.31}{SE}$.

5. Find the degrees of freedom using the Welch-Satterthwaite equation and determine the p-value.

Try solving on your own before revealing the answer!

Final Answer:

Test statistic $t \approx 2.54$, p-value $\approx 0.013$

Since p-value < 0.05, there is evidence the means are different for red and blue backgrounds.

Q11. Test scores from students who attended a study group: 88, 69, 96, 54, 75, 81, 83, 85. Not attended: 70, 98, 18, 53, 85, 66, 79. Test if there is a difference in variance using a p-value hypothesis test.

Background

Topic: F-Test for Equality of Variances

This question tests your ability to compare the variances of two independent samples using the F-distribution.

Key Terms and Formulas

• Null hypothesis: $H_0: \sigma_1^2 = \sigma_2^2$

• Alternative hypothesis: $H_1: \sigma_1^2 \neq \sigma_2^2$

• Test statistic: $F = \frac{s_1^2}{s_2^2}$ (larger variance in numerator)

• Degrees of freedom: $df_1 = n_1 - 1$, $df_2 = n_2 - 1$

Step-by-Step Guidance

1. Calculate the sample variances for both groups.

2. Compute the F-statistic: $F = \frac{\text{larger variance}}{\text{smaller variance}}$.

3. Find the p-value using the F-distribution with the appropriate degrees of freedom.

4. Compare the p-value to your significance level to decide whether to reject $H_0$.

Try solving on your own before revealing the answer!

Final Answer:

F-statistic $\approx 2.13$, p-value $\approx 0.32$

Since p-value > 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude the variances are different.

Q12. A coach tests a new protein supplement on 7 runners, recording race times with and without the supplement. Using a 90% significance level:

• a. Construct a confidence interval for the difference in means and interpret it.

• b. Use a hypothesis test to conclude if the supplement results in lower times.

Background

Topic: Paired t-Test for Means (Dependent Samples)

This question tests your ability to analyze paired data (before-and-after measurements) using confidence intervals and hypothesis testing.

Key Terms and Formulas

• Difference for each pair: $d_i = x_{i,\text{without}} - x_{i,\text{with}}$

• Mean difference: $\bar{d}$

• Standard deviation of differences: $s_d$

• Confidence interval: $\bar{d} \pm t^* \cdot \frac{s_d}{\sqrt{n}}$

• Hypothesis test: $H_0: \mu_d = 0$, $H_1: \mu_d > 0$ (if testing for lower times with supplement)

• Test statistic: $t = \frac{\bar{d}}{s_d/\sqrt{n}}$

Step-by-Step Guidance

1. Calculate the difference in times for each runner (without - with supplement).

2. Compute the mean and standard deviation of the differences.

3. Find the critical t-value for 90% confidence and $n-1$ degrees of freedom ($n=7$).

4. Construct the confidence interval: $\bar{d} \pm t^* \cdot \frac{s_d}{\sqrt{n}}$.

5. For the hypothesis test, calculate the t-statistic and compare to the critical value or find the p-value.

Try solving on your own before revealing the answer!

Final Answer:

90% CI: (0.01, 0.41) minutes

Since the interval is above zero, the supplement may reduce race times.

Test statistic $t \approx 2.09$, p-value $\approx 0.04$

Since p-value < 0.10, there is evidence the supplement lowers race times.