Question

A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You know from the water bottling company specifications that the standard deviation of the amount of water per bottle is 0.03 gallon. You select a random sample of 100 bottles, the mean amount of water per 1-gallon bottle is 0.994 gallon.

a. Is there evidence that the mean amount is different from 1.0 gallon? (Use ? = 0.01)

b. Compute the p-value and interpret its meaning.

c. Construct a 95% confidence interval estimate of the population mean amount of water per bottle.

d. Compare the results of (a) and (c). What conclusions do you reach?

Answer 1

(a) = 1, = 0.994, = 0.03, n = 100, = 0.01

The Hypothesis:

H0: = 1

Ha: 1

This is a 2 tailed test

The Test Statistic:

The test statistic is given by the equation:

The Critical Value:   The critical value (2 Tail) at = 0.01, Zcritical= +2.576 and -2.576

The Decision Rule: If Zobservedis >Zcritical or if Zobserved is < -Zcritical, Then reject H0.

The Decision: SinceZ falls in between +2.576 and -2.576, We Fail to Reject H0.

The Conclusion: There is insufficient evidence at the 99% significance level to conclude that the mean amount is different from 1 gallon.

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(b) The p Value: The p value (2 Tail) for Z= -2,is; p value = 0.0456

The Interpretation: The p value is the probability of getting a test statistic as extreme as or greater than the one obtained, assuming the null hypothesis is true.

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(c) The 99% CI

The Confidence Interval is given by ME, where

ME = Zcritical * \frac{s}{\sqrt{n}} = 2.576 * \frac{0.03}{\sqrt{100}} = 0.008

The Lower Limit = 994 - 0.008 = 0.986

The Upper Limit = 994 + 0.008 = 1.002

The 99% Confidence Interval is (0.986 , 1.002)

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(d) Based on the results of (a) and (c)

From (a) : We failed to reject H0 as per the hypothesis. From (c), we see that the confidence interval contains the value = 1, which means the null hypothesis H0: = 1, could be true and hence we fail to reject the null hypothesis.

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