# Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 randomly selected employees has a mean age of 22.4 years with a standard deviation of 3.8 years. Round to the nearest tenth.A] Determine the critical value ta/2 with n-the 1 degrees of freedomB] Determine the lower and upper bound of the confidence intervalC] Interpret the confidence interval.

Question

Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 randomly selected employees has a mean age of 22.4 years with a standard deviation of 3.8 years. Round to the nearest tenth.

A] Determine the critical value ta/2 with n-the 1 degrees of freedom

B] Determine the lower and upper bound of the confidence interval

C] Interpret the confidence interval.

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Step 1

Solution:

Part (a): Finding the critical value tα/2 with n-1 degrees of freedom:

Given that, the sample size n =19 and sample mean, x-bar =22.4 with its corresponding sample standard deviation, s = 3.8.

Here the confidence interval is 99%.

Therefore, α = 0.01.

The degrees of freedom, n - 1 = 19 - 1 = 18,

Step 2

From the EXCEL, using the formula, =T.INV.2T (0.01,18), the critical value is 2.878.

Therefore, the critical value tα/2 with 18 degrees of freedom is 2.878.

Step 3

Part (b): Finding the upper and lower bound of the confidence interval:

The population standard deviation is unknown here. Therefore, z-test cannot be used here. In this situation student’s t-distribution can be used. As the sample size is large it can be used th...

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