MATH 225N Week 7 Assignment: Conducting a Hypothesis Test for Mean – Population Standard Deviation Known P-Value Approach

Question 1: Jacqueline, a golfer, has a sample driving distance mean of 202.0 yards from 12 drives. Jacqueline claims that her average driving distance is 223 yards, and she attributes the lower average to chance. At a 2% significance level, does the data provide sufficient evidence to conclude that Jacqueline’s mean driving distance is less than 223 yards?

Answer:

• Null Hypothesis (H0): μ = 223 yards
• Alternative Hypothesis (Ha): μ < 223 yards
• Significance Level (α): 0.02
• Sample Mean (x̄): 202.0 yards
• Sample Size (n): 12
• Population Standard Deviation (σ): [Not Provided, Assumed Known for Test]
• Test Statistic (z): Use the z-test formula for calculating the z-value.
• Decision Rule: Reject H0 if p-value < α.
• Conclusion: Depending on the computed p-value, either reject or fail to reject H0.

Question 2: What is the p-value of a two-tailed one-mean hypothesis test with a test statistic of z = -1.59?

Answer:

• The p-value for z = -1.59 in a two-tailed test is approximately 0.112.
• Since it is a two-tailed test, you would multiply the area in one tail by 2.
• Conclusion: The exact p-value is 0.112.

Question 3: What is the p-value of a two-tailed one-mean hypothesis test with a test statistic of z = -1.73?

Answer:

• The p-value for z = -1.73 is approximately 0.084.
• Conclusion: The exact p-value is 0.084.

Question 4: Timothy, a bowler, has a sample game score mean of 202.1 from 11 games. He claims that his average game score is 182, and the higher average can be attributed to chance. At a 5% significance level, does the data provide sufficient evidence to conclude that Timothy’s mean game score is greater than 182?

Answer:

• Null Hypothesis (H0): μ ≤ 182
• Alternative Hypothesis (Ha): μ > 182
• Significance Level (α): 0.05
• Test Statistic (z): [Calculated based on provided data]
• Decision Rule: Reject H0 if p-value < α.
• Conclusion: Depending on the computed p-value, either reject or fail to reject H0.

Question 5: Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary’s sample mean throw was 55.5 meters based on 12 throws. At a 1% significance level, does the data provide sufficient evidence to conclude that Mary’s mean throw is less than 61 meters?

Answer:

• Null Hypothesis (H0): μ = 61 meters
• Alternative Hypothesis (Ha): μ < 61 meters
• Significance Level (α): 0.01
• Test Statistic (z): -1.99
• P-Value: 0.0233
• Decision Rule: Do not reject H0 because p-value > α.
• Conclusion: There is not enough evidence to conclude that Mary’s mean throw is less than 61 meters.

Question 6: Christina, a javelin thrower, has a sample throw mean of 62.3 meters from 29 throws. She claims her average throw is 57 meters and attributes the higher average to chance. At a 2% significance level, does the data provide sufficient evidence to conclude that Christina’s mean throw is greater than 57 meters?

Answer:

• Null Hypothesis (H0): μ ≤ 57 meters
• Alternative Hypothesis (Ha): μ > 57 meters
• Significance Level (α): 0.02
• Test Statistic (z): [Calculated based on provided data]
• Decision Rule: Reject H0 if p-value < α.
• Conclusion: Depending on the computed p-value, either reject or fail to reject H0.

Question 7: What is the p-value of a two-tailed one-mean hypothesis test with a test statistic of z = 0.27?

Answer:

• The p-value for z = 0.27 in a two-tailed test is approximately 0.789.
• Conclusion: The exact p-value is 0.789.

Question 8: Curtis, a statistician, claims that the average salary of an employee in Yarmouth is no more than $55,000 per year. Gina, his colleague, disagrees and randomly selects 61 employees, finding a sample mean of $56,500 with a standard deviation of $3,750. Using the alternative hypothesis Ha: μ > 55,000, find the test statistic and the p-value.

Answer:

• Null Hypothesis (H0): μ ≤ $55,000
• Alternative Hypothesis (Ha): μ > $55,000
• Test Statistic (t): Calculated value based on provided data.
• P-Value: The computed p-value should be rounded to three decimal places.
• Conclusion: Depending on the computed p-value, either reject or fail to reject H0.

Question 9: Marty, a typist, claims that his average typing speed is 72 words per minute. During a practice session, Marty has a sample typing speed mean of 84 words per minute based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is greater than 72 words per minute?

Answer:

• Null Hypothesis (H0): μ ≤ 72 words per minute
• Alternative Hypothesis (Ha): μ > 72 words per minute
• Significance Level (α): 0.05
• Test Statistic (z): 2.1
• P-Value: 0.018
• Decision Rule: Reject H0 because p-value < α.
• Conclusion: There is sufficient evidence to conclude that Marty’s mean typing speed is greater than 72 words per minute.

Question 10: Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives. She claims that her average driving distance is 207 yards, and the low average can be attributed to chance. At the 1% significance level, does the data provide sufficient evidence to conclude that Kathryn’s mean driving distance is less than 207 yards?

Answer:

• Null Hypothesis (H0): μ = 207 yards
• Alternative Hypothesis (Ha): μ < 207 yards
• Significance Level (α): 0.01
• Test Statistic (z): -1.46
• P-Value: 0.0721
• Decision Rule: Do not reject H0 because p-value > α.
• Conclusion: There is not enough evidence to conclude that Kathryn’s mean driving distance is less than 207 yards.