A simple random sample with n = 58 provided a sample mean of 24.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. to

Accepted Solution

Answer: (23.53, 25.47)Step-by-step explanation:The confidence interval for population mean(when population standard deviation is unknown) is given by :-[tex]\overline{x} \pm\ t^*\dfrac{s}{\sqrt{n}}[/tex] , where [tex]\overline{x}[/tex] = sample meann = sample size.s = sample standard deviation.t* = Critical value.Given : [tex]\overline{x}=24.5[/tex] s= 4.4 Confidence level = 90% =0.090Significance level = [tex]\alpha=1-0.90=0.10[/tex]Sample size : n= 58Degree of freedom : df= n-1= 57Using t-distribution table , the critical value :[tex]t_{\alpha/2,\ df}= t_{0.05,\ 57}=1.6720[/tex]Then, the confidence interval will be :-[tex]24.5 \pm\ (1.6720)\dfrac{4.4}{\sqrt{58}}[/tex][tex]24.5 \pm\ (1.6720)\dfrac{4.4}{7.61577310586}[/tex][tex]24.5 \pm\ 0.965995165263\approx24.5\pm0.97\\\\=(24.5-0.97,\ 24.5+0.97)\\\\=(23.53,\ 25.47)[/tex]Hence, a 90% confidence interval for the population mean = (23.53, 25.47)