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# Chapter 8 - PowerPoint PPT Presentation

Hypothesis Test. Chapter 8 . Steps to a Hypothesis Test. Hypotheses Null Hypothesis (Ho) Alternative Hypothesis (Ha) Alpha Distribution (aka model) Test Statistics and P-value Decision Conclusion. Steps to a Hypothesis Test. Can remember the steps by the sentence:

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Chapter 8

• Hypotheses

• Null Hypothesis (Ho)

• Alternative Hypothesis (Ha)

• Alpha

• Distribution (aka model)

• Test Statistics and P-value

• Decision

• Conclusion

• Can remember the steps by the sentence:

“Happy Aunts Make The Darndest Cookies”

• An attorney claims that more than 25% of all lawyers advertise. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising. At α = 0.05, is there enough evidence to support the attorney’s claim?

• Build the Alternative Hypothesis (Ha) first.

• based on the claim you are testing (you get this from the words in the problem)

• Three choices

• Ha: parameter ≠ hypothesized value

• Ha: parameter < hypothesized value

• Ha: parameter > hypothesized value

• Build Null Hypothesis (Ho) next.

• opposite of the Ha (i.e. = , ≥ , ≤ )

• We need to know what parameter we are testing and which of the three choices for alternative hypothesis we are going to use.

• “An attorney claims that more than 25% of all lawyers advertise” tells us that this is a test for proportions so our parameter is p.

• “claims that more than 25%” tells us that

Ha: p > .25 and therefore Ho: p ≤ .25

• Alpha = α = significance level

• How much proof we are requiring in order to reject the null hypothesis.

• The complement of the confidence level that we learned in the last chapter

• Usually given to you in the problem, if not, you can choose.

• Most popular alphas: 0.05, 0.01, and 0.10

• “At α = 0.05” is given to us in the problem so we just copy α = 0.05

• The model is the distribution used for the parameter that you are testing. These are just the same as we used in the confidence intervals.

• p and μ (n ≥ 30) use the normal distribution

• μ (n < 30) uses the t-distribution

• uses the chi-squared distribution

• The model used for a proportion is the normal.

• You will have a different test statistic for each of the four different parameters that we have learned about.

• p :

• μ (n ≥ 30) :

• You will have a different test statistic for each of the four different parameters that we have learned about.

• μ (n < 30) :

• :

• This is the evidence (probability) that you will get off of your chart and then compare against your criteria (alpha).

• You will need to find the appropriate probability that goes with your Ha.

• > and < Ha’s are called one-tailed tests.

• ≠ Ha’s are called two-tailed tests.

• For z and χ2 you have to take the > probability X2

• The formula for a test statistic for proportions is:

• So, from our problem we need a proportion from a sample (p-hat), the proportion from our hypothesis (po), and a sample size (n).

• “A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising” tells us that

• p-hat = 63/200 or 0.315

• From our hypothesis we know

• po = 0.25 (which means that qo = 0.75)

• “sample of 200” tells us that

• n = 200

• So our test statistic and p-value are

• We have two choices for decision

• Reject Ho

• Do Not Reject Ho

• If our evidence (p-value) is less than α we REJECT Ho.

• If our evidence (p-value) is greater than α we DO NOT REJECT Ho.

• Our p-value is 0.0170 and our alpha is 0.05

• So, since our p-value is less than our alpha our decision is: REJECT Ho.

• Conclusions

• Reject Ho

• “There is enough evidence to suggest (Ha).”

• Do Not Reject

• “There is not enough evidence to suggest (Ha).”

• Our decision to was to reject Ho, so our conclusion is:

“There is enough evidence to suggest that p>0.25”

• Ho: p ≤ 0.25

Ha: p > 0.25

• α = 0.05

• Model: Normal

• z = 2.12 and p-value = 0.0170

• Reject Ho

• There is enough evidence to suggest that p>0.25.

A researcher reports that the average salary of assistant professors is more than \$42,000. A sample of 30 assistant professors has a mean of \$43,260. At α = 0.05, test the claim that assistant professors earn more than \$42,000 a year. The standard deviation of the population is \$5230.

• Hypotheses

• Ho: μ ≤ \$42,000

• Ha: μ > \$42,000 (given claim is “more than”)

• Alpha

• α = 0.05 (given)

• Model

• Normal (n ≥ 30 and it’s a mean)

• Test statistic and p-value:

• Decision

• 0.0934 > 0.05 (p-value > alpha)

• DO NOT REJECT Ho

• Conclusion

• We do not have evidence to suggest that

μ > \$42,000.

A physician claims that joggers’ maximal volume oxygen uptake is greater than the average of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per kilogram (ml/kg) and a standard deviation of 6 ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physicians claim at α = 0.05?

• Hypotheses

• Ho: μ ≤ 36.7

• Ha: μ > 36.7

• Alpha

• α = 0.05 (given)

• Model

• t(14)

• Test statistic and p-value:

• Decision

• (0.01,0.025) < 0.05 (p-value < alpha)

• REJECT Ho

• Conclusion

• There is evidence to suggest that μ > 36.7.

A researcher knows from past studies that the standard deviation of the time it takes to inspect a car is 16.8 minutes. A sample of 24 cars is selected and inspected. The standard deviation was 12.5 minutes. At α=0.05, can it be concluded that the standard deviation has changed?

• Hypotheses

• Ho: σ = 16.8

• Ha: σ≠ 16.8

• Alpha

• α = 0.05 (given)

• Model

• χ2(23)

• Test statistic and p-value:

• Decision

• (0.05,0.10) > 0.05 (p-value > alpha)

• DO NOT REJECT Ho

• Conclusion

• There is not enough evidence to suggest that

σ≠ 16.8.