Question

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas. She takes a random sample of 25 pizzas and records their mean and standard deviation as 16.10 inches and 1.8 inches, respectively. She subsequently computes a 95% confidence interval of the mean size of all pizzas as [15.36, 16.84]. 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 1.8 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:

n=50

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=16.10$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size  

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

We need to find the degrees of freedom given by:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=\pm 1.96$

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

Replacing into formula (b) we got:

$n=(\frac{1.96(1.8)}{0.5})^2 =49.79 \approx 50$

So the answer for this case would be n=50 rounded up to the nearest integer