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.

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

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

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.

yes

step-by-step explanation:

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!