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# (Solved) MATH225 Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard Deviation Known P Value Approach

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Question

What is theÂ p-value of aÂ two-tailedÂ one-mean hypothesis test, with a test statistic ofÂ z0=âˆ’1.73? (Do not round your answer; compute your answer using a value from the table below.)

zâˆ’1.8âˆ’1.7âˆ’1.6âˆ’1.5âˆ’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

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 isÂ 61Â meters. During a practice session, Mary has a sample throw mean ofÂ 55.5Â meters based onÂ 12Â throws. At theÂ 1%Â significance level, does the data provide sufficient evidence to conclude that Mary’s mean throw is less thanÂ 61Â meters? Accept or reject the hypothesis given the sample data below.

• H0:Î¼=61Â meters;Â Ha:Î¼<61Â meters
• Î±=0.01(significance level)
• z0=âˆ’1.99
• p=0.0233

• Reject the null hypothesis becauseÂ |âˆ’1.99|>0.01.
• Do not reject the null hypothesis becauseÂ |âˆ’1.99|>0.01.
• Reject the null hypothesis because theÂ p-valueÂ 0233is greater than the significance levelÂ Î±=0.01.
• Do not reject the null hypothesis because the value ofÂ zis negative.
• Do not reject the null hypothesis because theÂ p-valueÂ 0233is greater than the significance levelÂ Î±=0.01.

Question

Marty, a typist, claims that his average typing speed isÂ 72Â words per minute. During a practice session, Marty has a sample typing speed mean ofÂ 84Â words per minute based onÂ 12Â trials. At theÂ 5%Â significance level, does the data provide sufficient evidence to conclude that his mean typing speed is greater thanÂ 72Â words per minute? Accept or reject the hypothesis given the sample data below.

• H0:Î¼â‰¤72Â words per minute;Â Ha:Î¼>72Â words per minute
• Î±=0.05(significance level)
• z0=2.1
• p=0.018

• Do not reject the null hypothesis becauseÂ 1>0.05.
• Do not reject the null hypothesis because the value ofÂ zis positive.
• Reject the null hypothesis becauseÂ 75>0.05.
• Reject the null hypothesis because theÂ p-valueÂ 018is less than the significance levelÂ Î±=0.05.
• Do not reject the null hypothesis because theÂ p-valueÂ 018is less than the significance levelÂ Î±=0.05.

Question

Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more thanÂ \$55,000Â per year. Gina, his colleague, believes this to be incorrect, so she randomly selectsÂ 61Â employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to beÂ \$56,500Â per year with a sample standard deviation ofÂ 3,750. Using the alternative hypothesisÂ Ha:Î¼>55,000, find the test statisticÂ tÂ 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 theÂ p-value of aÂ two-tailedÂ one-mean hypothesis test, with a test statistic ofÂ z0=âˆ’1.59? (Do not round your answer; compute your answer using a value from the table below.)

zâˆ’1.8âˆ’1.7âˆ’1.6âˆ’1.5âˆ’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 isÂ 253Â yards. During a practice session, Nancy has a sample driving distance mean ofÂ 229.6Â yards based onÂ 18Â drives. At theÂ 2%Â significance level, does the data provide sufficient evidence to conclude that Nancy’s mean driving distance is less thanÂ 253Â yards? Accept or reject the hypothesis given the sample data below.

• H0:Î¼=253Â yards;Â Ha:Î¼<253Â yards
• Î±=0.02(significance level)
• z0=âˆ’0.75
• p=0.2266

• Do not reject the null hypothesis because theÂ p-valueÂ 0.2266Â is greater than the significance levelÂ Î±=0.02.
• Do not reject the null hypothesis because the value ofÂ zÂ is negative.
• Do not reject the null hypothesis becauseÂ |âˆ’0.75|>0.02.
• Reject the null hypothesis becauseÂ |âˆ’0.75|>0.02.
• Reject the null hypothesis because theÂ p-valueÂ 0.2266Â is greater than the significance levelÂ Î±=0.02.

Question

Kathryn, a golfer, has a sample driving distance mean ofÂ 187.3Â yards fromÂ 13Â drives. Kathryn still claims that her average driving distance isÂ 207Â yards, and the low average can be attributed to chance. At theÂ 1%Â significance level, does the data provide sufficient evidence to conclude that Kathryn’s mean driving distance is less thanÂ 207Â yards? Given the sample data below, accept or reject the hypothesis.

• H0:Î¼=207Â yards;Â Ha:Î¼<207Â yards
• Î±=0.01(significance level)
• z0=âˆ’1.46
• p=0.0721

• Reject the null hypothesis because theÂ p-valueÂ 0.0721Â is greater than the significance levelÂ Î±=0.01.
• Do not reject the null hypothesis because theÂ p-valueÂ 0.0721Â is greater than the significance levelÂ Î±=0.01.
• Reject the null hypothesis becauseÂ |âˆ’1.46|>0.01.
• Do not reject the null hypothesis becauseÂ |âˆ’1.46|>0.01.
• Do not reject the null hypothesis because the value ofÂ zÂ is negative.

Question

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

• H0:Î¼=140;Â Ha:Î¼<140
• Î±=0.05(significance level)
• z0=âˆ’0.52
• p=0.3015
• Do not reject the null hypothesis because theÂ p-valueÂ 3015is greater than the significance levelÂ Î±=0.05.

## Question

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

• H0:Î¼=143;Â Ha:Î¼<143
• Î±=0.05Â (significance level)
• z0=âˆ’1.07
• p=0.1423

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