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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. the restaurant wishes to test whether sales have increased as a result of the advertising campaign. if the level of significance is 0.05, what is the decision?

User Bill Odom
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Answer: The null and alternative hypothesis 6000 > 6000 This is an one tailed test Test statistic 6300 = 1000 n= 49 Thus , test statistic is = 2.1 Q1. At 0.025 le

Explanation:

User Fabian Silva
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