Question

A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 98% confidence interval for the mean and are assuming that the population standard deviation for the number of energy drinks consumed each week is 1.1. The study found that for a sample of 1027 teenagers the mean number of energy drinks consumed per week is 5.9. Construct the desired confidence interval. Round your answers to one decimal place.

Answer 1

The 98% confidence interval for the mean number of energy drinks consumed per week by teenagers is (5.8, 6).

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.98)/(2) = 0.01`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.01 = 0.99`, so Z = 2.327.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.327(1.1)/(√(1027)) = 0.1`

The lower end of the interval is the sample mean subtracted by M. So it is 5.9 - 0.1 = 5.8 drinks per week.

The upper end of the interval is the sample mean added to M. So it is 5.9 + 0.1 = 6 drinks per week.

The 98% confidence interval for the mean number of energy drinks consumed per week by teenagers is (5.8, 6).