Question

As a​ follow-up to a report on gas​ consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage for their vehicles. A simple random sample of 96 SUV owners was​ selected, and the owners were asked to report their highway mileage. The results that were summarized from the sample data were x overbarequals19.6 mpg and sequals5.6 mpg. Based on these sample​ data, compute and interpret a 95​% confidence interval estimate for the mean highway mileage for SUVs.

Answer 1

Answer:

The 95​% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).

Step-by-step explanation:

Our sample size is 96.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 96-1 = 95$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 95 and 0.025 in the t-distribution table, we have $T = 1.9855$.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{5.6}{\sqrt{96}} = 0.57$

Now, we multiply T and s

$M = T*s = 1.9855*0.57 = 1.31$

For the lower end of the interval, we subtract the mean by M. So $19.6 - 1.31 = 18.29$

For the upper end of the interval, we add the mean to M. So $19.6 + 1.31 = 20.91$

The 95​% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).