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

Answer:

She must take a sample of 56.

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

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

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

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

$0.5 = 1.96\frac{1.9}{\sqrt{n}}$

$0.5\sqrt{n} = 1.96*1.9$

$\sqrt{n} = \frac{1.96*1.9}{0.5}$

$(\sqrt{n})^2 = (\frac{1.96*1.9}{0.5})^2$

$n = 55.5$

Rounding up:

She must take a sample of 56.