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Chapter 12: Hypothesis Tests – Large Samples
Note:
This is a right-tailed test because the direction of the inequality sign
in the alternative hypothesis is to the right.
Example 12 - 5:
A publication reported that Kentucky’s per capita
consumption of alcohol is 1.81 gallons per year, ranking it in the bottom
10% nationally. A student polls 300 Kentuckians and finds a mean
consumption of 2.0 gallons per year, and naturally claims that Kentucky’s
consumption is higher than that reported. Test the student’s claim at the 5%
level of significance. Assume that the standard deviation of the
consumption is 0.30 gallons.
Summary information:
=300,
= 0.30 gal,
= 0.05,
̅
(sample mean) =
2.0 gallons,
= 1.645, and
= 1.81 gallons. Also,
√ ⁄
= 0.0173.
Since the student would like to establish that the average consumption is
higher, the alternative hypothesis should reflect this belief.
Solution:
1.81 gallons
1.81 gallons
:
̅
√ ⁄
= (2.0 – 1.81)/0.0173 = 10.9827
: For a significance level of
= 0.05, reject the null hypothesis if the
computed test statistic value
= 10.9827 >
= 1.645.
Conclusion
: Since 10.9827 > 1.645, reject
. There is sufficient sample
evidence to confirm the student’s claim. That is, there is sufficient sample
evidence to claim that the average consumption of alcohol in Kentucky is
greater than 1.81 gallons per year at the 5% level of significance.
Note:
This means that there is a statistical significant difference between the
sample mean and the postulated mean of 1.81 gallons per year.
Figure 12-11
displays the test statistic in relation to the rejection region.
Observe that the test statistic falls in the rejection region.