Lindsay knows from past experience that the standard

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minutes. Lindsay knows from past experience that the standard deviation of wait times is 5.0 minutes. Follow the steps in the previous section to determine if Lindsay’s changes have reduced average wait time to less than 20 minutes using alpha = 0.05.
Z Test of Hypothesis for the MeanData20Level of Significance0.05Population Standard Deviation15Sample Size45Sample Mean18.3Intermediate CalculationsStandard Error of the Mean2.2361-0.7603Lower-Tail TestLower Critical Value-1.6450.2235Do not reject the null hypothesisNull Hypothesis m=ZTest Statisticp-Value
Your Turn # 4H0: Mu =18, process is performing as it should H1: Mu not equal to 18, process is not performing as it shoTwo Tail problemHypothesised mean18Population Standard Deviation0.25Sample Size40sample mean18.06Alpha0.05Alpha/20.025standard error0.0395Z test statistic1.5179Z crtical, upper1.96Z crtical, lower-1.96Z test statistic < Z crticalDo not Reject the null p-value0.1290p-value > Alpha Do not Reject the null Kellogg’s has a process for filling boxes of Frosted Flakes cereal with 18 ounces of product when it is operating properly. Too much product in the box is wasteful to Kellogg’s, whereas too little product can lead to customer complaints. To determine if the filling process is performing as it should, a random sample of 40 boxes was examined, and the average weight of a box was found to be 18.06 ounces. Historically, the filling process has a standard deviation of 0.25 ounces. Follow the six steps for conducting a two-tail hypothesis test to determine if the filling process is operating properly, using alpha = 0.05.
ould hypothesishypothesis
Z Test of Hypothesis for the MeanData18Level of Significance0.05Population Standard Deviation15Sample Size40Sample Mean18.06Intermediate CalculationsStandard Error of the Mean2.37170.0253Two-Tail TestLower Critical Value-1.9600Upper Critical Value1.96000.9798Do not reject the null hypothesisNull Hypothesis m=ZTest Statisticp-Value

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