1. Estimating 1 – 2 and p1 – p2
Section 9.4

2. Objectives Distinguish between independent and dependent samples.
Compute confidence intervals for 1 – 2 when 1 and 2 are known.
Compute confidence intervals for 1 – 2 when 1 and 2 are unknown.
Compute confidence intervals for p1 – p2 using the normal approximation.

3. In this section, we will use samples from two populations to create confidence intervals for the difference between population parameters.

5. Example–Confidence interval for 1 – 2, 1 and 2 known
A random sample of n1 = 167 reports from the period before the fire showed that the average catch was x1 = 5.2 trout per day. Assume that the standard deviation of daily catch per fisherman during this period was 1 = 1.9 .
Another random sample of n2 = 125 fishing reports 5 years after the fire showed that the average catch per day was x2 = 6.8 trout. Assume that the standard deviation during this period was 2 = 2.3.

6. Example–Confidence interval for 1 – 2, 1 and 2 known
Compute a 95% confidence interval for, the difference of population means.
Solution:
n1 = n2 = 125
x1 = x2 = 6.8
1 = 2 = 2.3
z0.95 = 1.96

7. Example – Solution

8. Example – Solution The 95% confidence interval is
(x1 – x2) – E < 1 – 2 < (x1 – x2) + E
(5.2 – 6.8) – 0.50 < 1 – 2 < (5.2 – 6.8)
–2.10 < 1 – 2 < –1.10

9. Example – Confidence interval for 1 – 2, 1 and 2 known
Interpretation What is the meaning of the confidence interval computed in part (b)?
Solution:
We are 95% confident that the interval –2.10 to –1.10 fish per day is one of the intervals containing the population difference 1 – 2, where the population means represent the average daily catch before and after the fire.
In words, we are 95% sure that the average catch before the fire was less than the average catch after the fire.

10. When 1 and 2 are unknown, we turn to a Student’s t distribution.
Confidence interval for 1 – 2, When 1 and 2 Are unknown
When 1 and 2 are unknown, we turn to a Student’s t distribution.

12. Example – Confidence interval for 1 – 2, When 1 and 2 unknown
Suppose that a random sample of 29 college students was randomly divided into two groups. The first group of n1 = 15 people was given 1/2 liter of caffeine before going to sleep. The second group of n2 = 14 people was given no caffeine before going to sleep. Everyone in both groups went to sleep at 11 P.M. The average brain wave activity (4 to 6 A.M.) was determined for each individual in the groups.

13. Group 1 (x1 values): n1 = 15 (with caffeine) x1  19.65 and s1  1.86
Example – Confidence interval for 1 – 2, When 1 and 2 unknown
Assume the average brain wave distribution in each group is mound-shaped and symmetric.
Group 1 (x1 values): n1 = 15 (with caffeine)
Average brain wave activity in the hours 4 to 6 A.M.
x1  and s1  1.86
Group 2 (x2 values): n2 = 14 (no caffeine)
x2  6.59 and s2  1.91

14. For d.f. use the smaller of n1 – 1 and n2 – 1.
Example – Confidence interval for 1 – 2, When 1 and 2 unknown
Compute a 90% confidence interval for 1 – 2 , the difference of population means.
Solution:
Find t0.90.
For d.f. use the smaller of n1 – 1 and n2 – 1.
d.f. = n2 – 1 = 14 – 1 =13.
t0.90  1.771

15. Example – Solution The c confidence interval is
(x1 – x2) – E < 1 – 2 < (x1 – x2) + E
(19.65 – 6.59) – 1.24 < 1 – 2 < (19.65 – 6.59)
11.82 < 1 – 2 < 14.30

16. Example – Confidence interval for 1 – 2, When 1 and 2 unknown
Interpretation What is the meaning of the confidence interval you computed in part (b)?
Solution:
We are 90% confident that the interval between 11.8 and hertz is one that contains the difference between caffeine and no caffeine.

18. Example – Confidence interval for p1 – p2
Based on a study of over 650 people in a Sleep Lab, it was found that about one-third of all dream reports contain feelings of fear, anxiety, or aggression. There is a conjecture that if a person is in a good mood when going to sleep, the proportion of “bad” dreams (fear, anxiety, aggression) might be reduced. Suppose that two groups of subjects were randomly chosen for a sleep study.

19. Example – Confidence interval for p1 – p2
Group I, before going to sleep, the subjects spent 1 hour watching a comedy movie.
n1 = 175 dreams recorded
r1 = 49 dreams with feelings of anxiety, fear, or aggression
In group II, the subjects did not watch a movie but simply went to sleep. n2 = 180 dreams recorded
r2 = 63 dreams with feelings of anxiety, fear, or aggression

20. Example – Solution

21. Example – Confidence interval for p1 – p2
What is p1 – p2? Compute a 95% confidence interval for p1 – p2.
Solution:
To find a confidence interval for p1 – p2, we need the values of zc, and then E. z0.95 = 1.96.

22. Example 10(b) – Solution

23. Example 10(b) – Solution E = zc = 1.96(0.0492)  0.096
(0.28 – 0.35) – < p1 – p2 < (0.28 – 0.35)
–0.166 < p1 – p2 < 0.026

24. Example – Confidence interval for p1 – p2
Interpretation What is the meaning of the confidence interval constructed in part (b)?
Solution:
We are 95% sure that the interval between –16.6% and 2.6% is one that contains the percentage difference of “bad” dreams for group I and group II.
The comedy movies before bed helped some people reduce the percentage of “bad” dreams, but at the 95% confidence level, we cannot say that the population difference is reduced.

25. 9.4 Estimating 1 – 2 and p1 – p2 Summarize Notes Read section 9.4
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