Question

If Xˉ=93, S=8, and n=36, and assuming that the population is normally distributed, construct a 90% confidence interval estimate of the population mean, μ. Click here to view page 1 of the table of critical values for the 1 distribution. crick here to view page 2 of the lable of critical values for the t distribution. (Round to two decimal places as needed.)

Answer 1

The 90% confidence interval estimate for the population mean (μ) is (91.08, 94.92).

Step-by-step explanation:

To construct a confidence interval for the population mean (μ) when the population standard deviation is unknown, we use the t-distribution. Given the sample mean (X\bar), sample standard deviation (S), and sample size (n), we can calculate the margin of error and then construct the confidence interval.

The formula for the margin of error (E) is given by:

`\[E = t_(\alpha/2) * (S)/(√(n))\]`

In this formula:

`- \(t_(\alpha/2)\)` is the critical value from the t-distribution table for the desired confidence level and degrees of freedom (df = n - 1).

- Sis the sample standard deviation.

- n is the sample size.

For a 90% confidence interval with df = 35 from (n - 1), the critical value
`\(t_(\alpha/2)\)` is approximately 1.690. Substituting the given values:

`\[E = 1.690 * (8)/(√(36))\]`

Calculating this gives the margin of error, which is then used to construct the confidence interval:

`\[E \approx 1.690 * (8)/(6) \approx 2.25\]`

Finally, we construct the confidence interval by adding and subtracting the margin of error from the sample mean:

`\[CI = (\bar{X} - E, \bar{X} + E)\]`

`\[CI = (93 - 2.25, 93 + 2.25)\]`

`\[CI = (91.08, 94.92)\]`

Therefore, we can be 90% confident that the true population mean falls within the interval (91.08, 94.92).