Question

A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a standard deviation of 1.6 credit hours. Estimate the mean credit hours taken by a student each quarter using a 95% confidence interval. Round to the nearest thousandth.

Answer 1

Answer:  $15.3\pm0.198$

OR

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval$(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

, where n is the sample size

$s$ = sample standard deviation.

$\overline{x}$= Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$ .

Given : Confidence level = 95% = 0.95

Significance level = $\alpha=1-0.95=0.05$

$s= 1.6$

$\overline{x}=15.3$

sample size : n= 250 , which is extremely large ( than n=30) .

So we assume sample standard deviation is the population standard deviation.

thus , $\sigma=1.6$

By standard normal  distribution table ,

Two tailed z-value for Significance level of 0.05 :

$z_{\alpha/2}=z_{0.025}=1.96$

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

$15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)$

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)