Question

A 90% confidence interval for a population mean is (65, 77). the population distribution is approximately normal and the population standard deviation is unknown. this confidence interval is based on a simple random sample of 25 observations. calculate the sample mean, the margin of error, and the sample standard deviation.

Answer 1

Part A:

The sample mean is at the midpoint of the lower and the upper values of the confidence interval.

Thus, the sample mean is given by:

(77 + 65) / 2 = 71

Therefore, the sample mean is 71.



Part 2:

The margin of error is the distance between the upper and the lower confidence interval values from the sample mean.

Thus, margin of error is given by 77 - 71 = 71 - 65 = 6.

Therefore, margin of error is 6.



Part 3:

The margin of error is given by

$ME= t_{\alpha/2}\left(\frac{s}{\sqrt{n}}\right)$

where $t_{\alpha/2}$ is the value of the t-distribution given the degree of freedom; s is the sample standard deviation and n is the sample size.

Since the sample size is 25, the degree of freedom is 25 - 1 = 24.
$\frac{\alpha}{2} = \frac{(1-0.9)}{2} = \frac{0.1}{2} =0.05$

From the t-distrbution table, we have:

$t_{\alpha/2}=1.710882$

Thus, margin of error is given by:

$6=1.710882\left( \frac{s}{\sqrt{25}} \right) \\ \\ \Rightarrow\left( \frac{s}{\sqrt{25}} \right)= \frac{6}{1.710882} \\ \\ \Rightarrow\frac{s}{5}=3.506963 \\ \\ \Rightarrow s=5(3.506963)=\bold{17.53}$