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Step-by-Step Guidance for Hypothesis Testing and P-Value Interpretation (Chapter 8)

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Q1. The test statistic in a right-tailed test is z = 0.52. Find the P-value and state the conclusion about the null hypothesis at a 0.05 significance level.

Background

Topic: Hypothesis Testing for Proportions (Z-Test)

This question tests your ability to interpret a test statistic (z-score) in a right-tailed hypothesis test, find the corresponding P-value, and make a decision about the null hypothesis using a given significance level.

Key Terms and Formulas

  • P-value: The probability, assuming the null hypothesis is true, of obtaining a result at least as extreme as the observed test statistic.

  • Right-tailed test: The area to the right of the test statistic under the standard normal curve.

  • Significance level (\( \alpha \)): The threshold for rejecting the null hypothesis (here, \( \alpha = 0.05 \)).

Step-by-Step Guidance

  1. Identify the test statistic: \( z = 0.52 \).

  2. Since this is a right-tailed test, the P-value is the area to the right of \( z = 0.52 \) under the standard normal curve.

  3. Use the standard normal (Z) table to find the area to the left of \( z = 0.52 \), then subtract this value from 1 to get the right-tail area (the P-value).

  4. Compare the P-value to the significance level \( \alpha = 0.05 \) to decide whether to reject or fail to reject the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: 0.3015; fail to reject the null hypothesis

Using the Z-table, the area to the left of 0.52 is approximately 0.6985. The P-value is 1 - 0.6985 = 0.3015. Since 0.3015 > 0.05, we fail to reject the null hypothesis.

Q2. The test statistic in a two-tailed test is z = -1.63. Find the P-value and state the conclusion about the null hypothesis at a 0.05 significance level.

Background

Topic: Hypothesis Testing for Proportions (Two-Tailed Z-Test)

This question tests your ability to find the P-value for a two-tailed test and make a decision about the null hypothesis.

Key Terms and Formulas

  • Two-tailed test: The P-value is the combined area in both tails beyond the absolute value of the test statistic.

  • P-value for two-tailed test: \( P = 2 \times P(Z < z) \) if z is negative, or \( P = 2 \times P(Z > z) \) if z is positive.

Step-by-Step Guidance

  1. Identify the test statistic: \( z = -1.63 \).

  2. Find the area to the left of \( z = -1.63 \) using the Z-table.

  3. Multiply this area by 2 to account for both tails (since it's a two-tailed test).

  4. Compare the resulting P-value to \( \alpha = 0.05 \) to determine whether to reject the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: 0.1032; fail to reject the null hypothesis

The area to the left of -1.63 is about 0.0516. Multiply by 2: 0.1032. Since 0.1032 > 0.05, fail to reject the null hypothesis.

Q3. With H1: p ≠ 3/5, the test statistic is z = 0.78. Find the P-value and state the conclusion about the null hypothesis.

Background

Topic: Hypothesis Testing for Proportions (Two-Tailed Z-Test)

This question tests your ability to interpret a test statistic for a two-tailed test and determine the P-value and conclusion.

Key Terms and Formulas

  • Alternative hypothesis: \( H_1: p \neq 3/5 \) (two-tailed test).

  • P-value for two-tailed test: \( P = 2 \times P(Z > |z|) \).

Step-by-Step Guidance

  1. Identify the test statistic: \( z = 0.78 \).

  2. Find the area to the right of \( z = 0.78 \) using the Z-table.

  3. Multiply this area by 2 for the two-tailed test.

  4. Compare the P-value to the significance level to decide on the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: 0.4354; fail to reject the null hypothesis

The area to the right of 0.78 is about 0.2177. Multiply by 2: 0.4354. Since this is greater than typical significance levels, fail to reject the null hypothesis.

Q4. With H1: p < 2/3, the test statistic is z = -1.93. Find the P-value and state the conclusion about the null hypothesis.

Background

Topic: Hypothesis Testing for Proportions (Left-Tailed Z-Test)

This question tests your ability to find the P-value for a left-tailed test and make a decision about the null hypothesis.

Key Terms and Formulas

  • Left-tailed test: The P-value is the area to the left of the test statistic.

Step-by-Step Guidance

  1. Identify the test statistic: \( z = -1.93 \).

  2. Use the Z-table to find the area to the left of \( z = -1.93 \) (this is the P-value).

  3. Compare the P-value to the significance level to decide whether to reject the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: 0.0268; reject the null hypothesis

The area to the left of -1.93 is about 0.0268. Since this is less than 0.05, reject the null hypothesis.

Q5. An entomologist claims that fewer than 14 in ten thousand male fireflies are unable to produce light due to a genetic mutation. If the null hypothesis is rejected, state the conclusion in nontechnical terms.

Background

Topic: Interpreting Hypothesis Test Results

This question tests your ability to translate statistical conclusions into plain language, specifically relating to the original claim.

Key Terms and Concepts

  • Rejecting the null hypothesis: There is sufficient evidence to support the alternative hypothesis (the claim).

Step-by-Step Guidance

  1. Understand the claim: The entomologist claims the true proportion is less than 14 in ten thousand.

  2. Recognize that rejecting the null hypothesis means there is sufficient evidence to support the alternative hypothesis (the claim).

  3. Express this conclusion in nontechnical terms, directly addressing the original claim.

Try solving on your own before revealing the answer!

Final Answer: There is sufficient evidence to support the claim that the true proportion is less than 14 in ten thousand.

Rejecting the null hypothesis means the data supports the entomologist's claim.

Q6. According to a recent poll, 53% of Americans would vote for the incumbent president. In a random sample of 100 people, 45% would vote for the incumbent. Test the claim that the actual percentage is 53% using a 0.10 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and conclusion.

Background

Topic: Hypothesis Testing for a Population Proportion

This question tests your ability to set up and conduct a hypothesis test for a population proportion using sample data.

Key Terms and Formulas

  • Null hypothesis: \( H_0: p = 0.53 \)

  • Alternative hypothesis: \( H_1: p \neq 0.53 \) (two-tailed test)

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

  • P-value: The probability of observing a test statistic as extreme as the one calculated, under the null hypothesis.

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: p = 0.53 \), \( H_1: p \neq 0.53 \).

  2. Calculate the sample proportion: \( \hat{p} = 0.45 \).

  3. Compute the standard error: \( SE = \sqrt{\frac{0.53 \times 0.47}{100}} \).

  4. Calculate the test statistic: \( z = \frac{0.45 - 0.53}{SE} \).

  5. Find the P-value for the calculated z (two-tailed test: multiply the one-tail area by 2).

  6. Compare the P-value to \( \alpha = 0.10 \) to decide whether to reject the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: Fail to reject the null hypothesis; there is not sufficient evidence to warrant rejection of the claim that the actual percentage is 53%.

The calculated z is about -1.60, and the P-value is about 0.0548. Since 0.0548 > 0.10, fail to reject the null hypothesis.

Q7. The mean of 60 temperature measurements is 20°C, with a known standard deviation of 1.5°C. Test the claim that the population mean is 22°C using a 0.05 significance level.

Background

Topic: Hypothesis Testing for a Population Mean (Z-Test, Known σ)

This question tests your ability to conduct a hypothesis test for a population mean when the population standard deviation is known.

Key Terms and Formulas

  • Null hypothesis: \( H_0: \mu = 22 \)

  • Alternative hypothesis: \( H_1: \mu \neq 22 \) (two-tailed test)

  • Test statistic (z): \( z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \)

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: \mu = 22 \), \( H_1: \mu \neq 22 \).

  2. Calculate the standard error: \( SE = \frac{1.5}{\sqrt{60}} \).

  3. Compute the test statistic: \( z = \frac{20 - 22}{SE} \).

  4. Find the P-value for the calculated z (two-tailed test).

  5. Compare the P-value to \( \alpha = 0.05 \) to decide on the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: Reject the null hypothesis; there is sufficient evidence to warrant rejection of the claim that the population mean temperature is 22°C.

The calculated z is about -10.33, and the P-value is very small (0.0002), which is less than 0.05.

Q8. A sample of 54 employees has a mean weight of 183.9 lb. Assuming σ is known to be 121.2 lb, test the claim that the population mean weight is less than 200 lb using a 0.10 significance level.

Background

Topic: Hypothesis Testing for a Population Mean (Z-Test, Known σ, Left-Tailed)

This question tests your ability to conduct a left-tailed hypothesis test for a population mean with a known standard deviation.

Key Terms and Formulas

  • Null hypothesis: \( H_0: \mu = 200 \)

  • Alternative hypothesis: \( H_1: \mu < 200 \) (left-tailed test)

  • Test statistic (z): \( z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \)

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: \mu = 200 \), \( H_1: \mu < 200 \).

  2. Calculate the standard error: \( SE = \frac{121.2}{\sqrt{54}} \).

  3. Compute the test statistic: \( z = \frac{183.9 - 200}{SE} \).

  4. Find the P-value for the calculated z (left-tailed test).

  5. Compare the P-value to \( \alpha = 0.10 \) to decide on the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean is less than 200 pounds.

The calculated z is about -0.98, and the P-value is about 0.1635, which is greater than 0.10.

Q9. Test the claim that for the adult population of one town, the mean annual salary is $30,000. Sample data: n = 17, x̄ = $22,298, s = $14,200. Use a significance level of α = 0.05. Find the test statistic, P-value, and state the final conclusion.

Background

Topic: Hypothesis Testing for a Population Mean (t-Test, Unknown σ)

This question tests your ability to conduct a t-test for a population mean when the population standard deviation is unknown and the sample size is small.

Key Terms and Formulas

  • Null hypothesis: \( H_0: \mu = 30,000 \)

  • Alternative hypothesis: \( H_1: \mu \neq 30,000 \) (two-tailed test)

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

  • Degrees of freedom: \( df = n - 1 \)

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: \mu = 30,000 \), \( H_1: \mu \neq 30,000 \).

  2. Calculate the standard error: \( SE = \frac{14,200}{\sqrt{17}} \).

  3. Compute the test statistic: \( t = \frac{22,298 - 30,000}{SE} \).

  4. Find the P-value for the calculated t (two-tailed test, df = 16).

  5. Compare the P-value to \( \alpha = 0.05 \) to decide on the null hypothesis.

Try solving on your own before revealing the answer!

Final Answer: Reject the null hypothesis; there is sufficient evidence to warrant rejection of the claim that μ = $30,000.

The calculated t is about -2.236, and the P-value is about 0.0399, which is less than 0.05.

Q10. Use a significance level of α = 0.01 to test the claim that μ > 2.85. The sample data consist of 9 scores for which x̄ = 3.17 and s = 0.57.

Background

Topic: Hypothesis Testing for a Population Mean (t-Test, Right-Tailed)

This question tests your ability to conduct a right-tailed t-test for a population mean with a small sample and unknown population standard deviation.

Key Terms and Formulas

  • Null hypothesis: \( H_0: \mu = 2.85 \)

  • Alternative hypothesis: \( H_1: \mu > 2.85 \) (right-tailed test)

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

  • Degrees of freedom: \( df = n - 1 \)

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: \mu = 2.85 \), \( H_1: \mu > 2.85 \).

  2. Calculate the standard error: \( SE = \frac{0.57}{\sqrt{9}} \).

  3. Compute the test statistic: \( t = \frac{3.17 - 2.85}{SE} \).

  4. Find the critical value for t at α = 0.01, df = 8, and compare the test statistic to this value.

  5. Decide whether to reject or fail to reject the null hypothesis based on the comparison.

Try solving on your own before revealing the answer!

Final Answer: Fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the mean is greater than 2.85.

The calculated t is about 1.68, which does not exceed the critical value of 2.896 for α = 0.01.

Q11. A test of sobriety involves measuring the subject's motor skills. Twenty randomly selected sober subjects take the test and produce a mean score of 41.0 with a standard deviation of 3.7. At the 0.01 level of significance, test the claim that the true mean score for all sober subjects is equal to 35.0.

Background

Topic: Hypothesis Testing for a Population Mean (t-Test, Two-Tailed)

This question tests your ability to conduct a two-tailed t-test for a population mean with a small sample and unknown population standard deviation.

Key Terms and Formulas

  • Null hypothesis: \( H_0: \mu = 35.0 \)

  • Alternative hypothesis: \( H_1: \mu \neq 35.0 \) (two-tailed test)

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

  • Degrees of freedom: \( df = n - 1 \)

Step-by-Step Guidance

  1. State the hypotheses: \( H_0: \mu = 35.0 \), \( H_1: \mu \neq 35.0 \).

  2. Calculate the standard error: \( SE = \frac{3.7}{\sqrt{20}} \).

  3. Compute the test statistic: \( t = \frac{41.0 - 35.0}{SE} \).

  4. Find the critical values for t at α = 0.01, df = 19, and compare the test statistic to these values.

  5. Decide whether to reject or fail to reject the null hypothesis based on the comparison.

Try solving on your own before revealing the answer!

Final Answer: Reject the null hypothesis; there is sufficient evidence to warrant rejection of the claim that the mean is equal to 35.0.

The calculated t is about 7.252, which is much greater than the critical value of ±2.861 for α = 0.01.

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