Answer:
Step-by-step explanation:
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
