Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth andrecord their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standarddeviation of 3,750. Using the alternative hypothesis H1 : 55.0appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.Right-Tailed T-TableprobabilityDegrees ofFreedom540.00040.00140.00240.00340.00440.00540.006456575859603.562 3.135 2.943 2.816 2.719 2.641 2.5763.558 3.132 2.941 2.814 2.7172.640 2.5743.554 3.130 2.939 2.8122.716 2.638 2.5723.550 3.127 2.937 2.810 2.714 2.636 2.5713.547 3.125 2.935 2.8082.712 2.635 2.5693.544 3.122 2.933 2.806 2.711 2.633 2.5683.540 3.120 2.931 2.805 2.709 2.632 2.567Provide your answer below, p-value -

Question
Asked Mar 19, 2019
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 H1 : 55.0
appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.
Right-Tailed T-Table
probability
Degrees of
Freedom
54
0.00040.00140.00240.00340.00440.00540.0064
56
57
58
59
60
3.562 3.135 2.943 2.816 2.719 2.641 2.576
3.558 3.132 2.941 2.814 2.7172.640 2.574
3.554 3.130 2.939 2.8122.716 2.638 2.572
3.550 3.127 2.937 2.810 2.714 2.636 2.571
3.547 3.125 2.935 2.8082.712 2.635 2.569
3.544 3.122 2.933 2.806 2.711 2.633 2.568
3.540 3.120 2.931 2.805 2.709 2.632 2.567
Provide your answer below
, p-value -
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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 H1 : 55.0 appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places. Right-Tailed T-Table probability Degrees of Freedom 54 0.00040.00140.00240.00340.00440.00540.0064 56 57 58 59 60 3.562 3.135 2.943 2.816 2.719 2.641 2.576 3.558 3.132 2.941 2.814 2.7172.640 2.574 3.554 3.130 2.939 2.8122.716 2.638 2.572 3.550 3.127 2.937 2.810 2.714 2.636 2.571 3.547 3.125 2.935 2.8082.712 2.635 2.569 3.544 3.122 2.933 2.806 2.711 2.633 2.568 3.540 3.120 2.931 2.805 2.709 2.632 2.567 Provide your answer below , p-value -

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check_circleExpert Solution
Step 1

State the appropriate hypothesis:

Denote the population mean as μ.

The investigator is interested to test whether the population mean salary of an employee in the city of Yarmouth is greater than $55,000 or not.

The hypotheses are given below:

Null hypothesis:

H0 : μ ≤ $55,000

That is, the population mean salary of an employee in the city of Yarmouth is less than or equal to $55,000.

Alternative hypothesis:

H1 : μ > $55,000 (Right tail test).

That is, the population mean salary of an employee in the city of Yarmouth is greater than $55,000.

Step 2

Obtain the test statistic value.

Here, the population standard deviation is not known.

Under null hypothesis the population mean salary of an employee in the city of Yarmouth is μ = $55,000.

The sample mean salary of an employee in the city of Yarmouth is x-bar = $56,500, the sample size is n = 61 and the sample standard deviation in the salary of an employee in the city of Yarmouth is s = $3,750.

The test statistic value is obtained as 3.12 from the calculation given below:

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Step 3

Obtain the P–value:

Here, the test statistic value is 3.12.

The degree of freedom is n – 1 = 61 – 1 = 60.

From the given Right-Tailed T-Table the P–value is obtained as 0.001 by using the procedure given below:

  • The objective is to find the value of P (z > 3.12).
  • The given tabulated values are corresponding to the right tailed values.
  • Locate the value 60 in the degrees of freedom.
  • Move towards the right along the ...

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