Question

Asked Apr 23, 2019

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 t _{α/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 t_{α/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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