Question 21 [t-interval] a random sample of size 18 is drawn from a population that is normally distributed. the sample mean is 58.5, and the sample standard deviation is found to be 11.5. determine a 95% confidence interval about population mean.a. [52.78,64.22]b. [53.78,63.22]c. [53.18,63.81]d. [54.04,62.96]

Answer:Option a. [52.78,64.22]Step-by-step explanation:We are given that a random sample of size 18 is drawn from a population that is normally distributed. Sample mean is 58.5, and the sample standard deviation is found to be 11.5 i.e., X bar = 58.5  and  s = 11.5 The pivotal quantity for calculating 95% confidence interval is;                [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex] So, 95% confidence interval about population mean is given by;P(-2.110 < [tex]t_1_7[/tex] < 2.110) = 0.95 P(-2.110 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.110) = 0.95 P(-2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95 P(X bar - 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95 95% confidence interval about [tex]\mu[/tex] = [ X bar - 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] , X bar + 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ]                                                    = [ 58.5 - 2.110 * [tex]\frac{11.5}{\sqrt{18} }[/tex] , 58.5 + 2.110 * [tex]\frac{11.5}{\sqrt{18} }[/tex] ]                                                    = [ 52.78 , 64.22 ]Therefore, 95% confidence interval about population mean is [52.78 , 64.22].