Question

A major tire manufacturer wishes to estimate the mean tread life in miles for one of its tires. It wishes to develop a confidence interval estimate that would have a maximum sampling error of 500 miles with 90 percent confidence. A pilot sample of n=50 tires showed a sample standard deviation equal to​ 4,000 miles. Based on this​ information, what is the required sample​ size?

Answer 1

Answer:

The required sample​ size is n=8,660.

Step-by-step explanation:

We have to calculate the minimum sample size that will give us a maximum margin of error of 500 miles, with a 90% confidence.

A pilot sample of n=50 give a sample standard deviation of 4,000 miles.

With the pilot sample we can calculate the population standard deviation as:

$\sigma_M=\sigma/\sqrt{n }\\\\\sigma=\sqrt{n}\sigma_M=\sqrt{50}*4,000=7.071*4,000=28,284$

The equation for the margin of error is:

$E=z\cdot \sigma/\sqrt{n}$

The z-value for a 90% confidence interval is z=1.645.

Then, we can estimate the sample size as:

$n=\left(\dfrac{z\cdot \sigma}{E}\right)^2=\left(\dfrac{1.645\cdot 28,284}{500}\right)^2=93.06^2=8,659.28\approx8,660$

Answer 2

The Z-value for a 90 percent confidence is 1.645

To find the sample size use the formula:

Sample size = Z-value x SD / error)^2

Using the provided information:

SD = 4,000 miles

Error = 500 miles

Sample size = ((1.645) (4000)/ 500)^2 = 173.18 = 174