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A sample of 27 employees for the Department of Health and Human Services has the following salaries, in thousands of dollars. Assuming normality, use Excel to find the 98% confidence interval for the true mean salary, in thousands of dollars. Round your answers to two decimal places and use increasing order.



SalarySalary
71
70
69
65
72
69
72
72
71
72
73
66
68
71
71
69
70
72
72
71
68
68
75
73
71
66
70

Respuesta :

Using the t-distribution, it is found that:

The 98% confidence interval for the true mean salary, in thousands of dollars, is: (69.14, 71.38).

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Before building the confidence interval, we use a calculator to find the sample mean and the sample standard deviation. They are:

[tex]\overline{x} = 70.26[/tex]

[tex]s = 2.35[/tex]

We have the standard deviation for the sample, thus, the t-distribution is used.

The number of degrees of freedom is the sample size subtracted by 1. So

df = 27 - 1 = 26

98% confidence interval

Now, we have to find a value of T, which is found looking at the two-tailed t table, with 26 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex], thus T = 2.479.

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.479\frac{2.35}{\sqrt{27}} = 1.12[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 70.26 - 1.12 = 69.14.

The upper end of the interval is the sample mean added to M. So it is 70.26 + 1.12 = 71.38.

The 98% confidence interval for the true mean salary, in thousands of dollars, is: (69.14, 71.38).

A similar problem is given at https://brainly.com/question/15872108

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