The makers of a soft drink want to identify the average age of its consumers. A sample of 16 consumers is taken. The average age in the sample was 22.5 years with a standard deviation of 5 years. a) Construct a 95% confidence interval for the true average age of the consumers. b) Construct an 80% confidence interval for the true average age of the consumers. c) Discuss why the 95% and 80% confidence intervals are different.

Assuming a normal distribution for the age of the consumers. We want to determine a 95% confidence interval for the true average age of the consumers.

Number of sample, n = 16

Mean, u = 22.5 years

Standard deviation, s = 5 years

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

22.5 ± 1.96 × 5/√16

= 22.5 +/- 1.96 × 1.25

= 22.5 +/- 2.45

The lower end of the confidence interval is 22.5 - 2.45 =20.05

The upper end of the confidence interval is 22.5 + 2.45 =24.95

For 80% confidence interval,

the corresponding z value is 1.28. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

22.5 ± 1.28 × 5/√16

= 22.5 +/- 1.28× 1.25

= 22.5 +/- 1.6

The lower end of the confidence interval is 22.5 - 1.6 =20.9

The upper end of the confidence interval is 22.5 + 1.6 =24.1

95% provides a wider interval than 80%. The wider the possible values of the true mean, the more confident we are. The lesser the possible values of the true mean, the less confident we are