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 0.025,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.

Question

Quizplus · Accepted Answer

The null hypothesis is that the mean daily sales have not increased after the advertising campaign, which means the population mean is still $6,000. The alternative hypothesis is that the mean daily sales have increased after the advertising campaign, which means the population mean is greater than $6,000.

We can use a one-sample t-test since the population standard deviation is given as $1,000 and the sample size is greater than 30. Using a significance level of 0.025 and 48 degrees of freedom (n-1), the critical value is 2.010.

The test statistic can be calculated as:

t = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
t = (6,300 - 6,000) / (1,000 / sqrt(49))
t = 3.06

Since the calculated t-value of 3.06 is greater than the critical value of 2.010, we reject the null hypothesis. Therefore, we conclude that the mean daily sales have increased after the advertising campaign and is higher than $6,000 per day.

Sales at a Fast-Food Restaurant Average $6,000 Per Day