The mean monthly expenditure on gasoline per household in Middletown is determined by selecting a random sample of 100 households. The sample mean is $128, with a sample standard deviation of $38, what is the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household in Middletown? a. $134.66 b. $162.20 c. $129.38 d. $131.42 e. $13287

Answer:a. $134.66Step-by-step explanation:1) Previous concepts A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]\bar X =128[/tex] represent the sample mean for the sample  [tex]\mu[/tex] population mean (variable of interest) s=38 represent the sample standard deviation n=100 represent the sample size  2) Calculate the confidence intervalSince the sample size is large enough n>30. The confidence interval for the mean is given by the following formula: [tex]\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1) Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]Now we have everything in order to replace into formula (1):[tex]128-1.64\frac{38}{\sqrt{100}}=121.768[/tex]    [tex]128+1.64\frac{38}{\sqrt{100}}=134.232[/tex] The closest value would be $134.66 and that would be the answer for this case.