Question

Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,400. From past history, the restaurant knew that its population standard deviation is about $1,000. The value of the test statistic is ______.

Answer 1

Answer:

The value of the test statistic is 2.8.

Step-by-step explanation:

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value expected value for the population mean, $\sigma$ is the standard deviation and n is the size of the sample.

Sales at a fast-food restaurant average $6,000 per day

This means that $\mu = 6000$

Sample of 49

This means that $n = 49$

The sample showed average daily sales of $6,400.

This means that $X = 6400$

Population standard deviation is about $1,000.

This means that $\sigma = 1000$

The value of the test statistic is

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{6400 - 6000}{\frac{1000}{\sqrt{49}}}$

$z = 2.8$

The value of the test statistic is 2.8.