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Chapter 12: Hypothesis Tests – Large Samples
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Note that we will use the sample standard deviation
s
as a point estimate for
unknown population standard deviation
. This is allowed becausewe have
a large sample size (
= 100).
Since the student would like to establish that the average engagement is less
than 6 hours, this will be a left-tailed test.
Solution:
6 hours
6 hours
T.S
:
̅
√ ⁄
= (5.8 – 6)/0.1 = -2.0000.
D.R
: For a significance level of
= 0.05, reject the null hypothesis if
the computed test statistic value
= -2.0000 < -
= -1.645.
Conclusion
: Since –2.0000 < -1.645, reject
. There is sufficient sample
evidence to support the claim that the average media engagement for these
children is less than 6 hours per day at the 5% level of significance.
Note:
There is a statistically significant difference between the sample
mean and the postulated value of the population mean of 6 hours; however, a
practical difference indeed may not exist, because the beneficial effects of 5
hours and 48 minutes of media engagement versus 6 hours are likely
equivalent in a practical sense.
Figure 12-14
displays the test statistic in relation to the rejection region.
Observe that the test statistic falls in the rejection region.