he inspection division of the Lee County Weights and Measure

he inspection division of the Lee County Weights and Measures Department is interested in determining whether the proper amount of soft drink has been placed in a 2-liter bottle at the local bottling plant of a large nationally known soft-drink company. The bottling plant has informed the inspection division that the standard deviation for 2-liter bottle is 0.05 liter. A random sample of one hundred 2-liter bottles obtained from this bottling plant indicates a sample mean of 1.99 liters.

1. Use the critical value of the test statistics to determine if there evidence that the mean amount in the bottles is different from 2.0 liters at the .02 level of significance.

2. Set up the 95% confidence interval estimate of the population mean amount in the bottles. What conclusion do you reach based on this interval?

Solution

Let mu be the population mean

The test hypothesis is

Ho: mu=2.0 (i.e. null hypothesis)

Ha: mu not equal to 2.0 (i.e. alternative hypothesis)

The test statistic is

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

=(1.99-2)/(0.05/sqrt(100))

=-2

It is a two-tailed test.

Given a=0.02, the critical values are Z(0.01)=-2.33 or 2.33 (from standard normal table)

The rejection regions are if Z<-2.33 or Z>2.33, we reject the null hypothesis.

Since Z=-2 is between -2.33 and 2.33, we do not reject the null hypothesis.

So we can not conclude that there is evidence that the mean amount in the bottles is different from 2.0 liters

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Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

xbar- Z*s/vn= 1.99-1.96*(0.05/sqrt(100)) =1.9802

So the upper bound is

xbar+ Z*s/vn= 1.99+1.96*(0.05/sqrt(100)) =1.9998

Since the interval does not include 2, we can concllude that there is evidence that the mean amount in the bottles is different from 2.0