Question

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 20-inch pizzas. She takes a random sample of 30 pizzas and records their mean and standard deviation as 20.50 inches and 2.10 inches, respectively. She subsequently computes the 90% confidence interval of the mean size of all pizzas as [19.87, 21.13]. 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 2.10 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch? (You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answer to the nearest whole number.)

Answer 1

The sample must be of at least 48 pizzas.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

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.

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

She needs a sample of at least n, in which is found when
`M = 0.5, \sigma = 2.1`

`M = z*(\sigma)/(√(n))`

`0.5 = 1.645*(2.1)/(√(n))`

`0.5√(n) = 2.1*1.645`

`√(n) = (2.1*1.645)/(0.5)`

`(√(n))^(2) = ((2.1*1.645)/(0.5))^(2)`

`n = 47.73`

Rounding up

The sample must be of at least 48 pizzas.