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 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.05, have sales increased as a result of the advertising campaign?

a) Fail to reject the null hypothesis.

b) Reject the null hypothesis and conclude the mean is higher than $6,000 per day.

c) Reject the null hypothesis and conclude the mean is lower than $6,000 per day.

d) Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.

Answer 1

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

H₀: µ = 6000 versus H_a: µ > 6000

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 6000

Xbar = 6300

σ = 1000

n = 49

α = 0.05

Critical value = 1.6449

(by using z-table or excel)

Z = (6300 - 6000)/[1000/sqrt(49)]

Z = 2.1000

P-value = 0.0179

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

b) Reject the null hypothesis and conclude that the mean is higher than $6,000 per day.