Question

The weights of running shoes are normally distributed with a population standard deviation of 3 ounces and an unknown population mean. If a random sample of 23 running shoes is taken and results in a sample mean of 18 ounces, find a 90% confidence interval for the population mean.

Answer 1

Answer: $(16.926\, 19.074)$

Step-by-step explanation:

Given : Level of significance : $1-\alpha:0.95$

Then , significance level : $\alpha: 1-0.95=0.05$

Since , sample size : $n=23$ , which is small sample (n<30) so the test applied here is a t-test.

Sample mean : $\overline{x}=18$

Standard deviation: $\sigma: 3$

Degree of freedom= $n-1=23-1=22$

Using t-distribution table , Critical value : $\text{t-score}=t_{n-1, \alpha/2}=1.717144$

The confidence interval for population mean is given by :-

$\overline{x}\pm t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$18\pm(1.717144)\dfrac{3}{\sqrt{23}}\approx18\pm 1.074=(16.926\, 19.074)$

Hence,  a 90% confidence interval for the population mean=$(16.926\, 19.074)$