Question

5. Kal Tire installs automobile tires on a first-come first-served basis. A random sample of 50 customers experienced an average wait time of 93.7 minutes. Assume that the standard deviation of total wait time for all customers is 20.6 minutes. Determine the margin of error for a 95% confidence interval for this sample.

Answer 1

5.7099 is the margin of error.

What is standard deviation ?

A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean.

Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

According to the question,

The standard sampling error of the sample mean is σₓ ,

The sampling distribution is: N (u,σₓ/n)

Therefore,

By using the standard deviation formula:

σₓ = √∑(xi -μ)/N

Where,

∑= population standard deviation

N= the size of the population

xi= each value from the population

μ =the population mean

So ,

σₓ = $\frac{20.6}{\sqrt{50} }$ = 2.913

Since, a = 1- 95% = 0.05

therefore $Z\frac{a}{2}$ = (1-0.005 x 2) = 1.959964

Here, the margin of error for a 95% confidence interval for this sample is given by:

$Z\frac{a}{2}$ * σₓ = 1.959964 x 2.913    

           = 5.7099

5.7099 is the margin of error for a 95% confidence interval for this sample.

Learn more about standard deviation here:

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