1. Ch. 9 examples

2. Summary of Hypothesis test steps
Null hypothesis H0, alternative hypothesis H1, and preset α
Test statistic and sampling distribution
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results
2-tailed ex
H0: µ= 100
H1: µ ≠ 100
α = 0.05
Left tail ex
H0: µ = 200
H1: µ < 200
Right tail ex
H0: µ = 50
H1: µ > 50

3. Should you use a 2 tail, or a right, or left tail test?
2-tailed ex
H0: µ= __
H1: µ ≠ __
Left tail ex
H0: µ = ___
H1: µ < ___
Right tail ex
H0: µ = __
H1: µ > __
Test whether the average in the bag of numbers is or isn’t 100.
Test if a drug had any effect on heartrate.
Test if a tutor helped the class do better on the next test.
Test if a drug improved elevated cholesterol.

4. Type I and Type II error

5. Probabilities associated with error

6. Example #1- numbers in a bag
Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation = Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

7. Ex #1- Hypothesis Test for numbers in a bag
α = 0.05
Z = =
P-value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

8. Ex #2– new sample mean for numbers in a bag
If the sample mean is 95, redo the test:
H0: µ = 100
H1: µ ≠ 100
α = 0.05
Z = =
P-value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

9. 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.

10. Ex #3- cholesterol- test
H0: µ (fill in the correct hypotheses here)
H1: µ
α = 0.05
Z = =
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

11. 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?

12. Ex #4: tutor Z = = P-value and/or critical value 4. Test conclusion
H0: µ (fill in the correct hypotheses here)
H1: µ 60
α = 0.05
Z = =
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

13. 9.2– 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

14. 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:

15. Ex #5 t test – placement scores
H0: µ fill in the correct hypothesis here
H1: µ 75
α = 0.05
t = =
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

16. 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:

17. Ex #6– t example H0: µ 80 (fill in the correct hypotheses here)
α = 0.05
t = =
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

18. 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:
41,000 42,500 39,000 39,999
43,000 43,550 44,200

19. Ex #7- salaries H0: µ 40000 (fill in the correct hypotheses here)
α = 0.05
t = =
P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

20. Traditional Critical Value Approach
Redo Example #1
Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation = Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

21. Ex#1 redone with CV Z = = CV 4. Test conclusion
α = 0.05
Z = = CV
4. Test conclusion
If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α
If p-value > α, then test value is not in RR, and we do not reject H0
5. Interpretation of test results

22. Ex #3 redone with CV
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.

23. Ex#3- 5 steps- done with CV
H1: µ (fill in)
α = 0.05
Z = = CV
4. Test conclusion
If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α
If p-value > α, then test value is not in RR, and we do not reject H0
5. Interpretation of test results

24. 9.3 Testing Proportion p
Recall confidence intervals for p:
± z

25. Hypothesis tests for proportions
Null hypothesis H0, alternative hypothesis H1, and preset α
2. Test statistic and sampling distribution
P-value and/or critical value
z= =
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results
2-tailed ex
H0: p= .5
H1: p ≠ .5
α = 0.05
Left tail ex
H0: p = .7
H1: p < .7
Right tail ex
H0: p = .2
H1: p > .2

26. 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.

27. Ex #8 H0: p=.75 (fill in hypothesis) H1: p α =
Null hypothesis H0, alternative hypothesis H1, and preset α
H0: p=.75 (fill in hypothesis)
H1: p
α =
Test statistic and sampling distribution
Z = =
3. P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results

28. 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.

29. Ex #9 H0: p (fill in hypothesis) H1: p α =
Null hypothesis H0, alternative hypothesis H1, and preset α
H0: p (fill in hypothesis)
H1: p
α =
Test statistic and sampling distribution
Z = =
3. P-value and/or critical value
4. Test conclusion
If p-value ≤ α, we reject H0 and say that the data are significant at level α
If p-value > α, we do not reject H0
5. Interpretation of test results