Question

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 90%90% confidence interval with an error of no more than 0.070.07. A consultant has informed them that a previous study found the mean to be 6.76.7 fast food meals per week and found the standard deviation to be 1.31.3. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

Given Information:

Mean of fast food meals per week = μ = 6.76

Standard deviation of fast food meals per week = σ =  1.31

Confidence level = 90%

Margin of error = 0.070

Required Information:

Sample size = n = ?

Answer:

Sample size ≈ 948

Step-by-step explanation:

We know that margin of error is given by

$MoE = z \cdot (\frac{\sigma}{\sqrt{n} } )$

Where n is the required sample size, z is the value of z-score corresponding to 90% confidence level and σ is the standard deviation.

Re-arranging the above equation for n yields,

$n = (\frac{z \cdot \sigma }{MoE})^{2}$

For 90% confidence level the corresponding z-score is 1.645

$n = (\frac{1.645 \cdot 1.31 }{0.070})^{2}\\n = (38.785)^{2}\\n = 947.71\\n = 948$

Therefore, a minimum sample size of 948 meals is required to ensure a margin of error no more than 0.070.