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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. 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 $6,300 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice

A. Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.
B. Fail to reject the null hypothesis.
C. Reject the null hypothesis and conclude the mean is lower than $6,000 per day.
D. Reject the null hypothesis and conclude the mean is higher than $6,000 per day.

Respuesta :

Answer:

Option B) Fail to reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $6,000

Sample mean, [tex]\bar{x}[/tex] = $6,300

Sample size, n = 49

Alpha, α = 0.01

Population standard deviation, σ = $1,000

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 6000\text{ dollars per week}\\H_A: \mu > 6000\text{ dollars per week}[/tex]

We use one-tailed(right) z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{6300 - 6000}{\frac{1000}{\sqrt{49}} } = 2.1[/tex]

Now, [tex]z_{critical} \text{ at 0.01 level of significance } = 2.33[/tex]

Since,  

[tex]z_{stat} < z_{critical}[/tex]

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that sales have not increased as a result of the advertising campaign

Option B) Fail to reject the null hypothesis.

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