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ex1,p114 Page 1 Test of Normal Hypotheses Concerning One Population Mean Input Data HoMean 3 SaMean 2.75 n 50 StdDev 1 Alpha 0.05 Calculated Value z -1.768 Test for Left-Tail LftCrt_zVal -1.645 In addition to showing your printout state the null and alternative hypotheses , and the results of the hypothesis test in terms of the question using complete sentences. Excel exercise 1, page 114: The Macburger restaurant chain claims that the mean waiting time of customers for service is normally distributed with a mean of 3 minutes and a standard deviation of 1 minute. The quality- assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we say that the mean waiting time is less than 3 minutes? (a left-tailed test)

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ex1,p114 Page 2 Conclusion Reject Ho p-value 0.0385 Null hypothesis: The population mean waiting time equals three minutes. (H 0 : m = 3) Alternative hypothesis: The population mean is less than three minutes. (H a : m < 3) -- "left tailed" There is enough evidence to support the conclusion that the mean waiting time is less than 3 minutes. The result we have observed (Sample mean =2.75 minutes) would occur fewer than 4 times in one hundred samples of this size if the null hypothesis were true. The p-value (0.0385) is the probability that the null hypothesis would be rejected if it were true. As the p-value is smaller than alpha, we will reject the null hypothesis.
ex2,p132 Page 3 Test of Normal Hypotheses Concerning One Population Mean Input Data HoMean 10 SaMean 9 n 50 StdDev 2.8 Alpha 0.05 Calculated Value z -2.525 Test for Left-Tail LftCrt_zVal -1.645 Conclusion Reject Ho p-value 0.0058 Excel exercise 2, page 132: A new weight watching

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