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 ______.

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Answer:

The value of the test statistic is 2.8.

Step-by-step explanation:

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value expected value for the population mean, [tex]\sigma[/tex] 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 [tex]\mu = 6000[/tex]

Sample of 49

This means that [tex]n = 49[/tex]

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

This means that [tex]X = 6400[/tex]

Population standard deviation is about $1,000.

This means that [tex]\sigma = 1000[/tex]

The value of the test statistic is

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{6400 - 6000}{\frac{1000}{\sqrt{49}}}[/tex]

[tex]z = 2.8[/tex]

The value of the test statistic is 2.8.