Question

The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 16. A random sample of 310 salespeople was taken and the mean number of cars sold annually was found to be 66. Find the 90% confidence interval estimate of the population mean. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.

Answer 1

Answer:

The 90% confidence interval is (64.0101, 68.9899).

Step-by-step explanation:

Our sample size is 310.

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

$df = 310-1 = 309$.

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

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 309 and 0.05 in the two-sided t-distribution table, we have $T = 1.65$

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

$s = \frac{16}{\sqrt{310}} = 1.2060$

Now, we multiply T and s

$M = 1.65*1.2060 = 1.9899$

Then

LCL is the mean subtracted by M. So:

$LCL = 66 - 1.9899 = 64.0101$

UCL is the mean added to M. So:

$LCL = 66 + 1.9899 = 68.9899$

The 90% confidence interval is (64.0101, 68.9899).