Question 976507: The waiting time for customers at MacBurger Restaurants follows a normal distribution with a mean of 3 minutes and a standard deviation of 1 minute. At Warren Road MacBurger, the quality-assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes? State the null hypothesis and the alternate hypothesis. State the decision rule. Compute the value of the test static. What is your decision regarding Ho? What is the p-value? Interpret it.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The waiting time for customers at MacBurger Restaurants follows a normal distribution with a mean of 3 minutes and a standard deviation of 1 minute.
At Warren Road MacBurger, the quality-assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes.
At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?
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State the null hypothesis and the alternate hypothesis.
Ho: u = >= 3
Ha: u < 3 (claim)
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State the decision rule.
invT(0.05,49) = -1.6766
Reject Ho if the test statistic is less than -1.6766
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Compute the value of the test static.
t(2.75) = (2.75-3)/[1/sqrt(50)] = -0.035
What is your decision regarding Ho?
Fail to reject Ho. The test results do not support the claim
that u < 3
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What is the p-value? Interpret it.
P(t < -1.6766 when df = 49) = tcdf(-100,-0.035) = 0.0486
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Since the p-value is less than 5%, reject Ho.
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Cheers,
Stan H.
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