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 records their annual salaries. 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. Degrees of Freedom

0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.0064
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

Test statistic t = 3.12

P-value = 0.001

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year (Gina's claim).

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=61.

The sample mean is M=56500.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3750.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

df=n-1=61-1=60

This test is a right-tailed test, with 60 degrees of freedom and t=3.12, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average salary of an employee in the city of Yarmouth is is significantly greater than $55,000 per year.

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844