1. Section 9.4
Day 3

2. Power
Power of a significance test is the probability of rejecting the null hypothesis.

3. Power
Power of a significance test is the probability of rejecting the null hypothesis.
Best way to get more power to reject a false null hypothesis is to _______ ____ ______.

4. Power
Power of a significance test is the probability of rejecting the null hypothesis.
Best way to get more power to reject a false null hypothesis is to increase sample sizes.

5. Power
If we have reason to believe the population standard deviations are about equal, make the sample sizes the same.

6. Power
If we have reason to believe the population standard deviations are about equal, make the sample sizes the same.
If we have reason to believe that one population’s standard deviation is larger than the other’s, allocate our resources so you take a larger sample from the population with the larger standard deviation.

7. Page 632, P31

8. Power
Which level has the greater spread?

9. Page 632, P31 The previous data indicate that the
measurements from the bottom of the river
have a greater spread, so ….

10. Page 632, P31 Because the previous data indicate that the
measurements from the bottom of the river
have a greater spread, getting more of the
new measurements at the bottom of the
river would give the test greater power.

11. Page 636, E60

12. Page 636, E60
What are the treatments here?

13. Page 636, E60
What are the treatments here?

14. Page 636, E60
What are the experimental units here?

15. Page 636, E60 The experimental units are the panels, not
the individual members.
So, the size of each treatment group is 8
because 8 panels were assigned to each
texture.

16. Page 636, E60
Name:

17. Page 636, E60
This is a two-sided significance test for the difference between two means.
Question asked “Is there a statistically significant difference in mean palatability score between the two texture levels?”

18. Page 636, E60
Check conditions:

19. Page 636, E60 Check conditions:
This is an experiment. Panel members were randomly assigned to the treatment groups as they were recruited.

20. Page 636, E60 List 1: scores for coarse texture
List 2: scores for fine texture

21. Page 636, E60
Check conditions: Show your graphs.

22. Page 636, E60
Check conditions: Looking at distribution of scores for each texture, both distributions are fairly symmetric with no outliers. It’s reasonable to assume each sample came from a normally distributed population.

23. Page 636, E60
State hypotheses.

24. Page 636, E60
State hypotheses.

25. Page 636, E60
Compute test statistic, find P-value, and draw sketch.

26. Page 636, E60 Compute test statistic, find P-value, and draw sketch.
2-SampTTest μ1: ≠ μ2
Inpt: Data Pooled: No
List1: L1 Calculate
List2: L2
Freq1: 1
Freq2: 1

27. Page 636, E60 Compute test statistic, find P-value, and draw sketch.

28. Page 636, E60
If all the panels sampled food with a coarse texture and all the panels sampled food with a fine texture, I would reject the null hypothesis because the P-value of is less than the significance level of 0.05.

29. Page 636, E60
If all the panels sampled food with a coarse texture and all the panels sampled food with a fine texture, I would reject the null hypothesis because the P-value of is less than the significance level of 0.05.
Thus, there is sufficient evidence to support the claim that there is a statistically significant difference in the mean palatability score between the coarse texture food and the fine texture food.

30. Page 634, E58

31. Page 634, E58
a) These are not independent random samples of men and women but an observational study conducted by looking at medical charts in one city.

32. Page 634, E58
a) These are not independent random samples of men and women but an observational study conducted by looking at medical charts in one city.
The data is fairly symmetrical so the underlying populations probably are approximately normal. There is one outlier for the males.

33. Page 634, E58
a) The populations of men and women are larger than 10 times their sample sizes (respectively 10(91) = 910 and 10(84) = 840)

34. Page 634, E58
Although conditions not met, we need to find the P-value for part b. Name of test we would use:

35. Page 634, E58
Although conditions not met, we need to find the P-value for part b. Name of test we would use: One-sided significance test for the difference of two means It has been speculated that the mean amount of calcium in the blood is higher in women than in men.

36. Page 634, E58
State hypotheses.
Ho:

37. Page 634, E58 State hypotheses.
Ho: μf = μm, where μf and μm are the mean calcium levels in the blood of women and men, respectively.

38. Page 634, E58 State hypotheses.
Ho: μf = μm, where μf and μm are the mean calcium levels in the blood of women and men, respectively.
Ha: μf > μm
It has been speculated that the mean amount of calcium in the blood is higher in women than in men.

39. E58 2-SampTTest Inpt: Stats μ1 > μ2 x1: 2.39690048 Pooled: No
Sx1: Calculate
n1: 84
x2:
Sx2:
n2: 91

40. E58 2-SampTTest Inpt: Stats μ1 > μ2 x1: 2.39690048 (F) Pooled: No
Sx1: stdDev Calculate
n1: 84
x2: (M) t =
Sx2: p =
n2: 91

41. Page 634, E58
t ; P-value
3.95

42. Page 634, E58
Interpret P-value: If these could be considered independent random samples of men and women and if there was no difference between the mean calcium levels in the blood of men and women, the probability of a t-statistic of 3.95 or larger with samples of 84 women and 91 men is

43. Page 634, E58
Interpret P-value: If these could be considered independent random samples of men and women and if there was no difference between the mean calcium levels in the blood of men and women, the probability of a t-statistic of 3.95 or larger with samples of 84 women and 91 men is However, because this was an observational study, these results are questionable.

44. Page 632, E53

45. Page 632, E53
a) The treatments, raised in long days or raised in short days, were randomly assigned to subjects.

46. Page 632, E53

47. Page 632, E53

48. Page 632, E53
a) Both distributions are moderately skewed, but neither has any outliers.
Since the distribution of the difference
reduces skewness, and the t-procedure is
robust against non-normality, the conditions
for inference are adequately met to proceed.

49. Page 632, E53
b. Interval?

50. Page 632, E53
b. ( , 9.46)

51. Page 632, E53 c. The difference between the mean enzyme
concentration of all eight hamsters had they
all been raised in short days and the mean
concentration had they all been raised in
long days.

52. Page 632, E53 d. Because 0 is not in the confidence
interval, and the treatments were
randomly assigned, Kelly has statistically
significant evidence that the difference in
enzyme concentrations between the two
groups of hamsters is due to the difference
in the amount of daylight.

53. Page 633, E54

54. Page 633, E54
Treatments randomly assigned.

55. Page 633, E54
Treatments randomly assigned.
Short
Long

56. Page 633, E54 Treatments randomly assigned.
Data for short days is strongly skewed right and sample size is very small, so not reasonable to assume data for short days comes from a normally distributed population.

57. Page 633, E54 You are 95% confident that the difference
between the mean enzyme concentration of
all eight hamsters had they all been raised in
short days and the mean concentration had
they all been raised in long days is in the
interval ( , ).

58. Page 633, E54 You are 95% confident that the difference
between the mean enzyme concentration of
all eight hamsters had they all been raised in
short days and the mean concentration had
they all been raised in long days is in the
interval ( , ).
Because 0 is in the CI, it’s plausible there is no difference in the means.

59. Page 633, E54 c) No. Because 0 is in the confidence
interval, Kelly does not have statistically
significant evidence that the two treatments
would result in different mean enzyme
concentrations.

60. Page 635, E59

61. Page 635, E59 a) No randomization is mentioned in
assigning treatments to experimental units.
(The experimental units here are the 16
dishes.) The distributions are not symmetric,
and the dish with 10 mg of resin is highly
skewed. Experimental group sizes of 8 will
not be enough to compensate for this
degree of skewness. Conditions are not met.

62. Page 635, E59

63. Page 635, E59
t ± 2.153; P-value ≈

64. Page 635, E59
Had the conditions been met, I would not reject the null hypothesis because the P-value of is larger than the significance level of 0.05.

65. Page 635, E59 There is insufficient evidence to support
the claim that there is a statistically
significant difference in the mean number of
termites surviving had all dishes received a
dose of 5 mg of resin and the mean number
of termites surviving had all dishes received
a dose of 10 mg of resin.

66. Page 635, E59 c) The main concerns are the lack of
randomization and the skewness of the distribution
of termites surviving in these rather small
experimental groups—especially in the 10-mg
group. The experimenter should repeat the
experiment, randomly assigning treatments to
the dishes, and using larger number of dishes to
compensate for the skewness.

67. Page 634, E57

68. Page 634, E57
Instructor believed that students tend to make larger errors (in absolute value) ….
Make the necessary changes to the data in the table so you can test the instructor’s belief.

69. Page 634, E57

70. Page 634, E57

71. E57

72. E57
A sample size of 15 is not large enough to compensate for this large amount of skewness if we were examining these data sets individually.

73. E57
A sample size of 15 is not large enough to compensate for this large amount of skewness if we were examining these data sets individually. However, because the difference of the means is a bit more lenient, this amount of skewness should be okay for inference.

74. E57

75. E57

76. E57
I reject the null hypothesis because the P-value of is less than the significance level of α = 0.05.

77. E57
I reject the null hypothesis because the P-value of is less than the significance level of α = 0.05.
There is sufficient evidence to support the instructor’s claim that students tend to make larger errors (in absolute value) when the line segments are vertical than when the line segments are horizontal.

78. Questions?