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

Step-by-step 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=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}\\\\\Rightarrow\ z=(6400-6000)/((1000)/(√(49)))=2.8`

Hence, the value of the test statistic is 2.8