Question

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 13-inch pizzas. She takes a random sample of 37 pizzas and records their mean and standard deviation as 13.50 inches and 1.90 inches, respectively. She subsequently computes the 95% confidence interval of the mean size of all pizzas as [12.89, 14.11]. However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 1.90 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch

Answer 1

She must take a sample of 56.

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 - 0.95)/(2) = 0.025`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error 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.

Standard deviation estimate of 1.90

This means that
`\sigma = 1.9`

Wow large a sample must she take if she wants the margin of error to be under 0.5 inch?

This is n for which M = 0.5. So

`M = z(\sigma)/(√(n))`

`0.5 = 1.96(1.9)/(√(n))`

`0.5√(n) = 1.96*1.9`

`√(n) = (1.96*1.9)/(0.5)`

`(√(n))^2 = ((1.96*1.9)/(0.5))^2`

`n = 55.5`

Rounding up:

She must take a sample of 56.