Question

A fast food restaurant executive wishes to know how many fast food meals teenagers 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 1.4 . The study found that for a sample of 502 teenagers the mean number of fast food meals consumed per week is 3.1 . Construct the desired confidence interval. Round your answers to one decimal place. Lower end point and Upper end point

Answer 1

Answer:

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).

Step-by-step explanation:

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

$\alpha = \frac{1-0.85}{2} = 0.075$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the length of the sample. So

$M = 1.44*\frac{1.4}{\sqrt{502}} = 0.09$

The lower end of the interval is the mean subtracted by M. So it is 3.1 - 0.09 = 3.01 meals.

The upper end of the interval is the mean added to M. So it is 3.1 + 0.09 = 3.19 meals.

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).