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:  [tex]15.3\pm0.198[/tex]OR(15.102, 15.498)Step-by-step explanation:The formula to find the confidence interval[tex](\mu)[/tex] is given by :-[tex]\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex], where n is the sample size [tex]s[/tex] = sample standard deviation.[tex]\overline{x}[/tex]= Sample mean[tex]z_{\alpha/2}[/tex] = Two tailed z-value for significance level of [tex]\alpha[/tex] .Given : Confidence level = 95% = 0.95Significance level = [tex]\alpha=1-0.95=0.05[/tex][tex]s= 1.6[/tex] [tex]\overline{x}=15.3[/tex]sample size : n= 250 , which is extremely large ( than n=30) .So we assume sample standard deviation is the population standard deviation.thus , [tex]\sigma=1.6[/tex]By standard normal  distribution table ,Two tailed z-value for Significance level of 0.05 : [tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-[tex]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)[/tex]Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)