A marketing consultant is hired by a major restaurant chain wishing to investigate the preferences and spending patterns of lunch customers. The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch. A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be $15.50. Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is o = $3.50. To address the CEO's conjecture, the marketing manager carried out a hypothesis test of : u = 13.50 vs. : u > 13.50 and obtained a P-value = 0.002. At a significance level of a = 0.05, this %3D %3! %3D %3D result:

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A marketing consultant is hired by a major restaurant chain wishing to investigate the
preferences and spending patterns of lunch customers. The CEO of the chain
hypothesized that the average customer spends at least $13.50 on lunch. A survey of
25 customers sampled at one of the restaurants found the average lunch bill per
customer to be = $15.50. Based on previous surveys, the restaurant informs the
marketing manager that the standard deviation is o = $3.50. To address the CEO's
conjecture, the marketing manager carried out a hypothesis test of : µ = 13.50 vs. : µ
> 13.50 and obtained a P-value = 0.002. At a significance level of a = 0.05, this
result:
O a. proves without a doubt that the average lunch bill exceeds $13.50.
O b. proves without a doubt that the average lunch bill does not exceed $13.50.
O c. provides evidence against the null hypothesis hypothesis in favor of the alternative
hypothesis.
O d. does not provide evidence against the null hypothesis in favor of the alternative
hypothesis.
Transcribed Image Text:A marketing consultant is hired by a major restaurant chain wishing to investigate the preferences and spending patterns of lunch customers. The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch. A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be = $15.50. Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is o = $3.50. To address the CEO's conjecture, the marketing manager carried out a hypothesis test of : µ = 13.50 vs. : µ > 13.50 and obtained a P-value = 0.002. At a significance level of a = 0.05, this result: O a. proves without a doubt that the average lunch bill exceeds $13.50. O b. proves without a doubt that the average lunch bill does not exceed $13.50. O c. provides evidence against the null hypothesis hypothesis in favor of the alternative hypothesis. O d. does not provide evidence against the null hypothesis in favor of the alternative hypothesis.
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