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 0.01, 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

Solution :

This is one tailed test .

The null and alternative hypothesis is ,

H0 :   = 6000

Ha : > 6000

Test statistic = z

= ( - ) / / n

= (6300 - 6000) / 1000 / 49

Test statistic = 2.1

P(z > 2.1) = 1 - P(z < 2.1) = 1 - 0.9821

P-value = 0.0179

= 0.01

P-value >

Fail to reject the null hypothesis.