1. Significance Test for a Mean
Section 9.2 Day 2
Significance Test for a Mean

2. Dealing with Outliers Sometimes your data will contain outliers.
How should you deal with these?

3. Dealing with Outliers So, do the analysis twice
There might be a good reason to remove the outliers.
So, do the analysis twice

4. Dealing with Outliers So, do the analysis twice once with the outliers
There might be a good reason to remove the outliers.
So, do the analysis twice
once with the outliers

5. Dealing with Outliers So, do the analysis twice once with the outliers
There might be a good reason to remove the outliers.
So, do the analysis twice
once with the outliers
once without them

6. Dealing with Outliers
If you are fortunate . . .

7. Dealing with Outliers
If you are fortunate, your analysis will lead you to the same conclusion both times either reject or do not reject the null hypothesis.

8. Dealing with Outliers
If you have a “split decision”, then . . .

9. Dealing with Outliers
If you have a “split decision”, then you’ll need more data.

10. Dealing with Outliers
If you have a “split decision”, then you’ll need more data.
You do not want your conclusion to depend on what you assume about one or two observations.

11. Page 597, P24
Name the test

12. Name of Test One-sided significance test of a mean
One-sided because “you are convinced that the room tends to be colder.”

13. Check Conditions

14. Check Conditions
1) Problem states the temperatures were taken on randomly selected days.

15. Check Conditions
2) The plot of the data is fairly symmetrical with no outliers, so it’s reasonable to assume the sample came from a normal population. (You must show a plot.)

16. Check Conditions
3) The population of days on which the temperatures could be taken is more than 10 times the sample size, (10x7 = 70).
So, all conditions met.

17. Write Hypotheses

18. Write Hypotheses
Ho: μ = 72o F, where μ is the mean temperature at my seat

19. Write Hypotheses
Ho: μ = 72o F, where μ is the mean temperature at my seat
Ha: μ < 72o F
(because “you are convinced that the room tends to be colder.”)

20. Computations

21. Computations
T-Test
< 72
t =
p = .0131

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

23. Conclusion 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 claim that the mean temperature at your desk is less than 72o F.

24. Page 587, D9

25. Page 587, D9
a) If s increases while everything else remains the same, the absolute value of the test statistic, t, will decrease.
Thus, the P-value will ________.

26. Page 587, D9
a) If s increases while everything else remains the same, the absolute value of the test statistic, t, will decrease.
Thus, the P-value will increase.

27. Page 587, D9

28. Page 587, D9
b) If the sample size increases but everything else remains the same, the absolute value of the test statistic, t, will increase.
Thus, the P-value will __________.

29. Page 587, D9
b) If the sample size increases but everything else remains the same, the absolute value of the test statistic, t, will increase.
Thus, the P-value will decrease.

30. Page 596, P12

31. Page 596, P12 The red curve is the standard normal distribution.
The standard normal distribution has less area in the tails.
The t-distribution is more spread out with heavier tails.

32. Page 599, E23

33. Page 599, E23
One-sided significance test for the mean because the consumer group says the average weight is less than 10 oz.

34. Page 599, E23
Randomness: problem states bags of chips randomly selected

35. Page 599, E23 Randomness: bags of chips randomly selected
Normality: company claims that the weight of all bags is normally distributed

36. Page 599, E23 Randomness: bags of chips randomly selected
Normality: company claims that the weight of all bags is normally distributed
Boxplot of data shows

37. Page 599, E23
Boxplot of data shows distribution is fairly symmetric with no outliers so reasonable to assume sample came from normally distributed population

38. Page 599, E23 All conditions met.
Population size: Population of bags of chips produced would be more than 150 bags which is 10 times the sample size.
All conditions met.

39. Page 599, E23
Ho: μ = 10 oz where μ is the mean weight of bags of Munchie’s Potato Chips

40. Page 599, E23
Ho: μ = 10 oz where μ is the mean weight of bags of Munchie’s Potato Chips
Ha: μ < 10

41. Page 599, E23 TTest Inpt: Data μo: 10 List: L1 Freq: 1 μ: < μo
Calculate

42. Page 599, E23
t =
P-value = .012

43. Page 599, E23 I reject the null hypothesis because the
P-value of is less than the significance
level of α = 0.05.

44. Page 599, E23 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 claim that the mean weight of bags of Munchie’s Potato Chips is less than 10 oz.

45. Page 597, P18

46. Page 597, P18 a)
Ho: = 1700, where is the mean SAT score of State University students

47. Page 597, P18 a)
Ho: = 1700, where is the mean SAT score of State University students
Ha:

48. Page 597, P18 b)
TTest
Inpt: Stats
: 1700
x : 1535
sx: 250
n: 9

49. Page 597, P18
b) t = ± 1.98, P-value = .083
10% level:

50. Page 597, P18
b) t = ± 1.98, p-value = .083
10% level: reject because .083 < .10
5% level:

51. Page 597, P18
b) t = ± 1.98, p-value = .083
10% level: reject because .083 < .10
5% level: do not reject because .083 > .05
1% level:

52. Page 597, P18
b) t = ± 1.98, p-value = .083
10% level: reject because .083 < .10
5% level: do not reject because .083 > .05
1% level: do not reject because .083 > .01

53. Questions?