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Sales at a fast-food restaurant average $6,000 per day. The restaurant decided t

ID: 3312056 • Letter: S

Question

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 0.05, have sales increased as a result of the advertising campaign? Multiple Choice Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is lower than $6,000 per day. Reject the null hypothesis and conclude that the mean is equal to $6,000 per day. Reject the null hypothesis and conclude the mean is higher than $6,000 per day.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 6000
Alternative hypothesis: < 6000

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 142.86
DF = n - 1

D.F = 48
t = (x - ) / SE

t = 2.09

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 2.09

Thus the P-value in this analysis is 0.9817.

Interpret results. Since the P-value (0.9817) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Fail to reject the null hypothesis.

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