Question

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 98% confidence interval with an error of no more than 0.08. A consultant has informed them that a previous study found the mean to be 6.6 fast food meals per week and found the standard deviation to be 0.7. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

Answer:

415

Step-by-step explanation:

Confidence Level = 98%

Z-value for this confidence level = z = 2.326

Margin of error = E = 0.08

Mean = u = 6.6

Standard deviation = $\sigma=0.7$

Required Sample Size = n = ?

The formula for margin of error is:

$E=z\frac{\sigma}{\sqrt{n}}$

Re-arranging the equation for n, and using the given values we get:

$n=(\frac{z\sigma}{E} )^{2}\\\\ n=(\frac{2.326 \times 0.7}{0.08} )\\\\ n=415$

Thus, the minimum sample size required to create the specified confidence interval is 415