Week 7 Assignment: Conducting a Hypothesis Test for Mean – Population Standard Deviation Known: P-Value Approach
Question
Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance level, does the data provide sufficient evidence to conclude that Mary’s mean throw is less than 61 meters? Accept or reject the hypothesis given the sample data below.
- H0:μ=61 meters; Ha:μ<61 meters
- α=0.01 (significance level)
- z0=−1.99
- p=0.0233
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.)
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27? (Do not round your answer; compute your answer using a value from the table below).
Question
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? Accept or reject the hypothesis given the sample data below.
- H0:μ≤72 words per minute; Ha:μ>72 words per minute
- α=0.05 (significance level)
- z0=2.1
- p=0.018
Question
Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ>55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.
Right-Tailed T-Table
| probability | 0.0004 | 0.0014 | 0.0024 | 0.0034 | 0.0044 | 0.0054 | 0.0064 |
| Degrees of Freedom | |||||||
| 54 | 3.562 | 3.135 | 2.943 | 2.816 | 2.719 | 2.641 | 2.576 |
| 55 | 3.558 | 3.132 | 2.941 | 2.814 | 2.717 | 2.640 | 2.574 |
| 56 | 3.554 | 3.130 | 2.939 | 2.812 | 2.716 | 2.638 | 2.572 |
| 57 | 3.550 | 3.127 | 2.937 | 2.810 | 2.714 | 2.636 | 2.571 |
| 58 | 3.547 | 3.125 | 2.935 | 2.808 | 2.712 | 2.635 | 2.569 |
| 59 | 3.544 | 3.122 | 2.933 | 2.806 | 2.711 | 2.633 | 2.568 |
| 60 | 3.540 | 3.120 | 2.931 | 2.805 | 2.709 | 2.632 | 2.567 |
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.59? (Do not round your answer; compute your answer using a value from the table below.)
Question
Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session, Nancy has a sample driving distance mean of 229.6 yards based on 18 drives. At the 2% significance level, does the data provide sufficient evidence to conclude that Nancy’s mean driving distance is less than 253 yards? Accept or reject the hypothesis given the sample data below.
- H0:μ=253 yards; Ha:μ<253 yards
- α=0.02 (significance level)
- z0=−0.75
- p=0.2266
Question
Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives. Kathryn still 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? Given the sample data below, accept or reject the hypothesis.
- H0:μ=207 yards; Ha:μ<207 yards
- α=0.01 (significance level)
- z0=−1.46
- p=0.0721
Question
Christina, a javelin thrower, has a sample throw mean of 62.3 meters from 29 throws. Christina still claims that her average throw is 57 meters, and the high average can be attributed to chance. At the 2% significance level, does the data provide sufficient evidence to conclude that Christina’s mean throw is greater than 57 meters? Given the sample data below, accept or reject the hypothesis.
- H0:μ=57 meters; Ha:μ>57 meters
- α=0.02 (significance level)
- z0=2.61
- p=0.0045
Question
Timothy, a bowler, has a sample game score mean of 202.1 from 11 games. Timothy still claims that his average game score is 182, and the high average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Timothy’s mean game score is greater than 182? Given the sample data below, accept or reject the hypothesis.
- H0:μ=182; Ha:μ>182
- α=0.05 (significance level)
- z0=1.57
- p=0.0582
Question
Shawn, a competitor in cup stacking, has a sample stacking time mean of 9.2 seconds from 13 trials. Shawn still claims that her average stacking time is 8.5 seconds, and the high average can be attributed to chance. At the 4% significance level, does the data provide sufficient evidence to conclude that Shawn’s mean stacking time is greater than 8.5 seconds? Given the sample data below, accept or reject the hypothesis.
- H0:μ=8.5 seconds; Ha:μ>8.5 seconds
- α=0.04 (significance level)
- z0=0.61
p=0.2709
Solution
Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance level, does the data provide sufficient evidence to conclude that Mary’s mean throw is less than 61 meters? Accept or reject the hypothesis given the sample data below.
- H0:μ=61 meters; Ha:μ<61 meters
- α=0.01 (significance level)
- z0=−1.99
- p=0.0233
Answer: Do not reject the null hypothesis because the p-value
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? Accept or reject the hypothesis given the sample data below.
- H0:μ≤72 words per minute; Ha:μ>72 words per minute
- α=0.05 (significance level)
- z0=2.1
- p=0.018
Answer: Reject the null hypothesis because the p-value 0.018 is…………….please follow the link below to purchase the solution at $5
