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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

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JeanaShupp

Answer: a. 2.8

Explanation:

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

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

Sample mean = [tex]\overline{x}=\$6,400\text{ per day}[/tex]

Standard deviation : [tex]\sigma= \$1,000[/tex]

The test statistic for population mean is given by :-

[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\Rightarrow\ z=\dfrac{6400-6000}{\dfrac{1000}{\sqrt{49}}}=2.8[/tex]

Hence, the value of the test statistic is 2.8

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