A marketing consultant is hired by a major restaurant chain wishing to investigate the preference lunch customers. The CEO of the chain hypothesized that the average customer spends at least $ of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be previous surveys, the restaurant informs the marketing manager that the standard deviation is o = Normally distributed. To address the CEO's conjecture, the marketing manager carried out a hyp Họ : µ = 13.50 vs. Ha : µ > 13.50 and obtained a P-value = 0.077. At a significance level of a = 0.05, this 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 \( \bar{x} = \$14.50 \). Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is \( \sigma = \$3.50 \). Lunch bills are Normally distributed. To address the CEO’s conjecture, the marketing manager carried out a hypothesis test of

\[ H_0 : \mu = 13.50 \text{ vs. } H_a : \mu > 13.50 \]

and obtained a \( P \)-value = 0.077.

At a significance level of \( \alpha = 0.05 \), this result:

- ○ proves without a doubt that the average lunch bill exceeds $13.50.
- ○ proves without a doubt that the average lunch bill does not exceed $13.50.
- ○ provides evidence against the alternative hypothesis in favor of the null hypothesis.
- ● 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 \( \bar{x} = \$14.50 \). Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is \( \sigma = \$3.50 \). Lunch bills are Normally distributed. To address the CEO’s conjecture, the marketing manager carried out a hypothesis test of \[ H_0 : \mu = 13.50 \text{ vs. } H_a : \mu > 13.50 \] and obtained a \( P \)-value = 0.077. At a significance level of \( \alpha = 0.05 \), this result: - ○ proves without a doubt that the average lunch bill exceeds $13.50. - ○ proves without a doubt that the average lunch bill does not exceed $13.50. - ○ provides evidence against the alternative hypothesis in favor of the null hypothesis. - ● does not provide evidence against the null hypothesis in favor of the alternative hypothesis.
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