Question

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 85% confidence interval for the mean and are assuming that the population standard deviation for the number of fast food meals consumed each week is 1.3. The study found that for a sample of 548 adults the mean number of fast food meals consumed per week is 5.6. Construct the desired confidence interval. Round your answers to one decimal place.

Answer 1

The 85% confidence interval for the mean is between 5.5 and 5.7 fast food meals consumed per week.

Explanation:

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

`\alpha = (1-0.85)/(2) = 0.075`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.075 = 0.925`, so
`z = 1.44`

Now, find 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 = 1.44*(1.3)/(√(548)) = 0.08`

The lower end of the interval is the sample mean subtracted by M. So it is 5.6 - 0.08 = 5.52 = 5.5.

The upper end of the interval is the sample mean added to M. So it is 5.6 + 0.08 = 5.7.

The 85% confidence interval for the mean is between 5.5 and 5.7 fast food meals consumed per week.