# (solved) Math 225N Week 7 Assignment Conducting a Hypothesis for Mean-Population Standard Deviation Known P Value Approach

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## Math 225N Week 7 Assignment Conducting a Hypothesis for Mean-Population Standard Deviation Known P Value Approach

**Week 7 Assignment Conducting a Hypothesis for Mean-Population Standard Deviation Known P Value Approach**

**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.)

z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.0380.0460.0570.0690.090.0290.0370.0460.0560.068

Provide your answer 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.)

z0.10.20.30.40.50.000.5400.5790.6180.6550.6910.010.5440.5830.6220.6590.6950.020.5480.5870.6260.6630.6980.030.5520.5910.6290.6660.7020.040.5560.5950.6330.6700.7050.050.5600.5990.6370.6740.7090.060.5640.6030.6410.6770.7120.070.5670.6060.6440.6810.7160.080.5710.6100.6480.6840.7190.090.5750.6140.6520.6880.722

**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
- Select the correct answer below:

- Reject the null hypothesis because |−1.99|>0.01.
- Do not reject the null hypothesis because |−1.99|>0.01.
- Reject the null hypothesis because the p-value 0233is greater than the significance level α=0.01.
- Do not reject the null hypothesis because the value of zis negative.
- Do not reject the null hypothesis because the p-value 0233is greater than the significance level α=0.01.

**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

Select the correct answer below:

- Do not reject the null hypothesis because 1>0.05.
- Do not reject the null hypothesis because the value of zis positive.
- Reject the null hypothesis because 75>0.05.
- Reject the null hypothesis because the p-value 018is less than the significance level α=0.05.
- Do not reject the null hypothesis because the p-value 018is less than the significance level α=0.05.

**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.)

z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.0380.0460.0570.0690.090.0290.0370.0460.0560.068

**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

Select the correct answer below:

Do not reject the null hypothesis because the p-value 0.2266 is greater than the significance level α=0.02.

Do not reject the null hypothesis because the value of z is negative.

Do not reject the null hypothesis because |−0.75|>0.02.

Reject the null hypothesis because |−0.75|>0.02.

Reject the null hypothesis because the p-value 0.2266 is greater than the significance level α=0.02.

**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

Select the correct answer below:

Reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01.

Do not reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01.

Reject the null hypothesis because |−1.46|>0.01.

Do not reject the null hypothesis because |−1.46|>0.01.

Do not reject the null hypothesis because the value of z is negative.

**Question**

Ruby, a bowler, has a sample game score mean of 125.8 from 25 games. Ruby still claims that her average game score is 140, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Ruby’s mean game score is less than 140? Given the sample data below, accept or reject the hypothesis.

## Question

Marie, a bowler, has a sample game score mean of 129.2 from 24 games. Marie still claims that her average game score is 143, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Marie’s mean game score is less than 143? Given the sample data below, accept or reject the hypothesis.