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 

.𝟔𝟏𝟕𝟎

Practice 

.𝟎𝟎𝟓𝟓

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%.