Question

the gas mileages (in miles per gallon) of 24 randomly selected sports cars are listed in the accompanying table. assume the mileages are not normally distributed. use the standard normal distribution or the t - distribution to construct a 99% confidence interval for the population mean. justify your decision. if neither distribution can be used, explain why. interpret the results. click the icon to view the sports car gas mileages. let σ be the population standard deviation and let n be the sample size. which distribution should be used to construct the confidence interval? neither distribution can be used to construct the confidence interval, since the population is not normally distributed and n < 30. identify the confidence interval. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. ( ) (round to one decimal place as needed.) b. neither the standard normal distribution nor the t - distribution can be used to construct the interval.

Explanation:

To construct a confidence interval using the standard normal distribution ($z$-distribution) or the $t$-distribution, certain conditions must be met. For the $z$-distribution, we need to know the population standard deviation $\sigma$ and either the population is normally distributed or the sample size $n\geq30$. For the $t$-distribution, the population does not need to have a known $\sigma$, but the population should be normally distributed (or approximately normally distributed) or $n\geq30$.

In this problem, we are told that the population (gas mileages of sports - cars) is not normally distributed and the sample size $n = 24<30$. Since neither the normality condition (for small - sample cases without transformation) nor the large - sample ($n\geq30$) condition is met for either the $z$ or $t$ distributions, we cannot use these distributions to construct the confidence interval.

Answer:

B. Neither the standard normal distribution nor the t - distribution can be used to construct the interval.