1. Ch. 10 examples Part 1 – z and t Ma260notes_Sull_ch10_HypTestEx.pptx

2. Test on a single population mean Intro: So far we’ve studied estimation, or confidence intervals, focusing on the middle 90% or 95% or 99% Now the focus is on Hypothesis Testing (using the 1% or 5% or 10% in the tails instead) As in Ch. 8 (CLT) and 9 (CI s), we’ll study both: – Means and – proportions

3. Courtroom Analogy- Two Hypotheses: Guilty or Not Guilty Our conclusion (the sentence) Verdict is Not guiltyVerdict is Guilty RealityDefendant is innocent Acquittal- Correct conclusion Error Defendant is guilty ErrorConviction- Correct conclusion

4. Hypothesis Testing in Statistics— Note the chance of two different errors (Type I, II)

5. Types of Hypotheses in Statistical Testing The null hypothesis H 0 is one which expresses the current state of nature of belief about a population – Note: The hypothesis with the equal sign is the null The alternative (or research) hypothesis (H 1 or H a ) is one which reflects the researcher’s belief. (It will always disagree with the null hypothesis). – Note: the alternative hypothesis can be one or two tailed. – In one tailed tests, the alternative is usually the claim

6. Summary of Hypothesis test steps 1.Null hypothesis H 0, alternative hypothesis H 1, and preset α 2.Test statistic and sampling distribution 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results 2-tailed ex H 0 : µ= 100 H 1 : µ ≠ 100 α = 0.05 Left tail ex H 0 : µ = 200 H 1 : µ < 200 α = 0.05 Right tail ex H 0 : µ = 50 H 1 : µ > 50 α = 0.05 CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

7. Should you use a 2 tail, or a right, or left tail test? Test whether the average in the bag of numbers is or isn’t 100. Test if a drug had any effect on heart rate. Test if a tutor helped the class do better on the next test. Test if a drug improved elevated cholesterol. 2-tailed ex H 0 : µ= __ H 1 : µ ≠ __ Left tail ex H 0 : µ = ___ H 1 : µ < ___ Right tail ex H 0 : µ = __ H 1 : µ > __

8. Example #1- numbers in a bag Recall the experiments with the bags of numbers. I claim that µ = 100. We’ll assume  =21.9. Test this hypothesis (that µ = 100 )if your sample size n= 20 and your sample mean was 90.

9. Ex #1- Hyp Test- numbers in a bag 1.H 0 : µ = 100 H 1 : µ ≠ 100 α = 0.05 2.Z = = 3.CV 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

10. Ex #2– new sample mean If the sample mean is 95, redo the test: 1.H 0 : µ = 100 H 1 : µ ≠ 100 α = 0.05 2.Z = = 3.CV 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

11. Ex #3: Left tail test- cholesterol A group has a mean cholesterol of 220. The data is normally distributed with σ= 15 After a new drug is used, test the claim that it lowers cholesterol. Data: n=30, sample mean= = 214.

12. Ex #3- cholesterol – left test 1.H 0 : µ 220 (fill in the correct hypotheses here) H 1 : µ 220 α = 0.05 2.Z = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

13. Ex #4- right tail- tutor Scores in a MATH117 class have been normally distributed, with a mean of 60 all semester. The teacher believes that a tutor would help. After a few weeks with the tutor, a sample of 35 students’ scores is taken. The sample mean is now 62. Assume a population standard deviation of 5. Has the tutor had a positive effect?

14. Ex #4: tutor 1.H 0 : µ 60 H 1 : µ 60 α = 0.05 2.Z = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

15. When  is unknown…use t tests Just like with confidence intervals, if we do not know the population standard deviation, we – substitute it with s (the sample standard deviation) and – Run a t test instead of a z test First, we’ll review using the t table

16. Critical values for t Find the CV for RIGHT tail examples when: – n=10,  =.05 – n=15,  =.10 Find the CV for LEFT tail examples when: – n=10,  =.05 – n=15,  =.10 Find the CV for TWO tail examples when: – n=10,  =.05 – n=15,  =.10 – n=20,  =.01

17. Ex #5– t test – placement scores The placement director states that the average placement score is 75. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

18. Ex #5 t test – placement scores 1.H 0 : µ 75 H 1 : µ 75 α = 0.05 2.t = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results 6.Confidence Interval CV approachP value approachdecision If test stat is in RRIf p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

19. Ex #6- placement scores The head of the tutoring department claims that the average placement score is below 80. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

20. Note: P values for t-tests We can use the t-table to approximate these values Use “Stat Crunch” on the online-hw to get more exact answers. – See the rectangle to the right of the data

21. Ex #6– t example 1.H 0 : µ 80(fill in the correct hypotheses here) H 1 : µ 80 α = 0.05 2.t = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RR p-value > αdo not reject H 0

22. Ex #7- salaries– t A national study shows that nurses earn $40,000. A career director claims that salaries in her town are higher than the national average. A sample provides the following data: 41,00042,50039,00039,999 43,00043,55044,200

23. Ex #7- salaries 1.H 0 : µ 40000(fill in the correct hypotheses here) H 1 : µ 40000 α = 0.05 2.t = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RR p-value > αdo not reject H 0

24. The Hypothesis Testing template Although hw may be worded differently, questions on the test will require you to do a 5 or 6 step answer (using either p value or CV) 1.Null and alternative hypotheses, H 0 and H 1, and preset α 2. Test statistic z = or t = 3.P-value and/or critical value 4.Test conclusion– Reject H 0 or not If test value is in RR (p-value ≤ α), we reject H 0. If test value is not in RR (p-value > α), we do not reject H 0 5.Interpretation of test results 6.Confidence Interval (for 2 tailed problems)

25. Ch.10 – part two Testing Proportion p Recall confidence intervals for p: ± z

26. Hypothesis tests for proportions 1.Null hypothesis H 0, alternative hypothesis H 1, and preset α 2.Sampling distribution Test statistic z = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results 2-tailed ex H 0 : p=.5 H 1 : p ≠.5 α = 0.05 Left tail ex H 0 : p =.7 H 1 : p <.7 α = 0.05 Right tail ex H 0 : p =.2 H 1 : p >.2 α = 0.05 CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

27. Ex #8- proportion who like job The HR director at a large corporation estimates that 75% of employees enjoy their jobs. From a sample of 200 people, 142 answer that they do. Test the HR director’s claim.

28. Ex #8 1.Null hypothesis H 0, alternative hypothesis H 1, and preset α H 0 : p=.75 H 1 : p α = 2.Test statistic and sampling distribution Z = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results 6.Confidence interval CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0

29. Ex #9 Previous studies show that 29% of eligible voters vote in the mid-terms. News pundits estimate that turnout will be lower than usual. A random sample of 800 adults reveals that 200 planned to vote in the mid- term elections. At the 1% level, test the news pundits’ predictions.

30. Ex #9 1.Null hypothesis H 0, alternative hypothesis H 1, and preset α H 0 : p (fill in hypothesis) H 1 : p α = 2.Test statistic and sampling distribution Z = = 3.P-value and/or critical value 4.Test conclusion 5.Interpretation of test results 6.Confidence Interval CV approachP value approachdecision If test stat is in RR (Rejection Region) If p-value ≤ αreject H 0 If test stat is not in RRp-value > αdo not reject H 0