Basics of Hypothesis
Testing8.2 Day 2
Homework Answers
ExampleConsider the Microsort example. The null
hypothesis is that p = 0.5. The sample that is returned shows that 13 out of 14 children whose parents used the product were girls.
Find the test statistic that would be used to challenge the null hypothesis.
The P-Value Method
P-Values
ExampleGiven that our test statistic is z = 3.21 for the
Microsort problem (remember that the alternative hypothesis is p > 0.5). What is the P-value?
Practice
Practice
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Practice
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The P-Value Method
When to RejectBy setting confidence level, we are determining
how unusual a statistic needs to be for us to consider rejecting the null hypothesis.
Ex. If we set the confidence level to 95%, we are saying that if we get back something that based on the assumptions of the null hypothesis would have a probability of it or something more extreme occurring of less than 5%, than the null hypothesis must not have been true.
In simpler terms, if the P-value is less than , we must reject the null hypothesis. If not,
we fail to reject.
ExampleThe P-value of our Microsort test was .0007.
Assuming our was set to be .05, would we be able to reject the null hypothesis?
The P-Value Method
ConclusionsYou can either reject the null hypothesis, or fail to
reject it. (which suggests that the alternative hypothesis is true)There is no way to prove the null hypothesis.
You must state your conclusion such that someone with no background in statistics could understand.There is sufficient evidence to reject the claim that
parents who use Microsort will have girls 50% of the time.
The sample data support the claim that Microsort improves the probability of having a baby girl.
Side Note: Confidence Interval Method
If you can create a confidence interval using the sample statistic, and it does not contain the value for the parameter given by the null hypothesis, you can reject the null hypothesis.
Homeworkp.410:
P-Value/Reject: #33-36
Conclusion: #38-39
Homework P.409: P-Value/Reject: 33-36
33. 0.0060; reject the null hypothesis.
34. 0.7264; fail to reject the null hypothesis
35. 0.0107; reject the null hypothesis
36. 0.0016; reject the null hypothesis
Conclusion: 38-39
38. There is sufficient evidence to support the claim that the percentage of on-time U.S. airlines flights is less than 75%.
39. There is not sufficient evidence to warrant rejection of the claim that the percentage of Americans who know their credit score is equal to 20%.
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