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

Answer:

The sample must be of at least 48 pizzas.

Step-by-step explanation:

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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 $M = 0.5, \sigma = 2.1$

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.645*\frac{2.1}{\sqrt{n}}$

$0.5\sqrt{n} = 2.1*1.645$

$\sqrt{n} = \frac{2.1*1.645}{0.5}$

$(\sqrt{n})^{2} = (\frac{2.1*1.645}{0.5})^{2}$

$n = 47.73$

Rounding up

The sample must be of at least 48 pizzas.