Suppose that the amount of cereal dispensed into a box is no

Suppose that the amount of cereal dispensed into a box is normally distributed. If the mean amount dispensed in a box is \"too small,\" then the proportion of \"underfilled\" boxes (boxes with less tlian 16 ounces of cereal in them) is too large. However. if the mean is \"too large,\" then the company loses money \"overfilling\" the boxes. The CEO of the company that makes Captain Crisp cereal, Mr. Statman, is concerned that the machines that dispense cereal into boxes do not have the proper (\"optimal\") setting for the mean amount dispensed. A random sample of 196 boxes was obtained, the sample mean amount of cereal in these 196 boxes was 16.07 ounces, the sample standard deviation was 0.21 ounces. Mr. Statman wants to test whether the mean amount dispensed is 1.61 ounces (which he considers to be the \"optimal\" value for the mean) or not. Perform the appropriate test using a 10% level of significance. Find the p-value of the test in (b).

Solution

Let mu be the population mean

The test hypothesis:

Ho: mu=16.1 (i.e. null hypothesis)

Ha: mu not equal to 16.1 (i.e. alternative hypothesis)

The test statistic is

Z=(xbar-mu)/(s/vn)

=(16.07-16.1)/(0.21/sqrt(196))

=-2

It is a two-tailed test.

Given a=0.1, the critical values are Z(0.05) = -1.645 or 1.645 (from standard normal table)

The rejection regions are if Z<-1.645 or Z>1.645, we reject the null hypothesis.

Since Z=-2 is less than -1.645, we reject the null hypothesis.

So we can not conclude that the mean amount dispensed is 16.1 ounces

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The p-value= 2*P(Z<-2) =0.0455 (from standard normal table)