Question

A lawsuit has been filed on behalf of workers at a warehouse who are required to undergo security searches at the end of the work day but are not paid for the time it takes to be searched. A random sample of 43 screenings (taken across multiple days) was used to estimate the mean screening time (in minutes), this resulted in a 95% confidence interval of (12.29, 16.17). (a) What is the margin of error of this sample? (2 decimals) (b) Suppose the researchers increased their confidence level to 99%, keeping everything else the same. How would this affect the following quantities? The standard error would V The critical value would The width of the confidence interval would V (c) Determine the sample size required to obtain a margin of error of 2.52 at the 99% confidence level. (You can assume the standard deviation is 6.30.)

Answer 1

(a) The margin of error is half the width of the confidence interval, which can be calculated as follows:

216.17−12.29​=1.94

So, the margin of error of this sample is 1.94.

(b) If the researchers increased their confidence level to 99%, keeping everything else the same, the following would happen:

The standard error would remain the same because it depends on the sample size and standard deviation, which are not changing.

The critical value would increase because a higher confidence level requires a larger critical value to capture more of the area under the normal curve.

The width of the confidence interval would increase because both the margin of error and the critical value are increasing.

© To determine the sample size required to obtain a margin of error of 2.52 at the 99% confidence level, we can use the formula for the margin of error for a mean:

E=zα/2​n​σ​

where E is the margin of error, zα/2​ is the critical value for a given confidence level, σ is the population standard deviation, and n is the sample size. Solving for n, we get:

n=(zα/2​Eσ​)2

Substituting in the values we have, with z0.005​=2.58 (from a standard normal table), σ=6.30, and E=2.52, we get:

n=(2.582.526.30​)2≈26

So, a sample size of 26 would be required to obtain a margin of error of 2.52 at the 99% confidence level.