Question

A population is known to be normally distributed with a standard deviation of 2.8. (a) Compute the 95% confidence interval on the mean based on the following sample of nine: 8, 9, 10, 13, 14, 16, 17, 20, 21. (b) Now compute the 99% confidence interval using the same data.

Answer 1

Explanation:

Mean = sum of terms/number of terms

The mean of the given sample is

(8 + 9 + 10 + 13 + 14 + 16 + 17 + 20 + 21)/9 = 14.2

a)

For a confidence level of 95%, the corresponding z value is 1.96.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

Where

n represents the number of samples

It becomes

14.2 ± 1.96 × 2.8/√9

= 14.2 ± 1.96 × 0.933

= 14.2 ± 1.83

The lower end of the confidence interval is 12.37

The upper end of the confidence interval is 16.03

b)

For a confidence level of 99%, the corresponding z value is 2.58

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

Where

n represents the number of samples

It becomes

14.2 ± 2.58 × 2.8/√9

= 14.2 ± 2.58 × 0.933

= 14.2 ± 2.4

The lower end of the confidence interval is 11.8

The upper end of the confidence interval is 16.6