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 days of sales were taken. They found that the average daily sales were $6,400 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. The value of the test statistic is _______.
a. 2.8 b. 1.96 c. 6,400 d. 6,000

Answer 1

Answer: a. 2.8

Explanation:

Given : Population mean : $\mu=\ 6,000\text{ per day}$

Sample size : n= 49> 30 , the sample is a large sample  we use z-test.

Sample mean = $\overline{x}=\ 6,400\text{ per day}$

Standard deviation : $\sigma= \ 1,000$

The test statistic for population mean is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\Rightarrow\ z=\dfrac{6400-6000}{\dfrac{1000}{\sqrt{49}}}=2.8$

Hence, the value of the test statistic is 2.8