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

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.

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

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.

Example–Confidence interval for 1 – 2, 1 and 2 known Compute a 95% confidence interval for, the difference of population means. Solution: n1 = 167 n2 = 125 x1 = 5.2 x2 = 6.8 1 = 1.9 2 = 2.3 z0.95 = 1.96

Example – Solution

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) + 0.50 –2.10 < 1 – 2 < –1.10

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.

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.

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.

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  19.65 and s1  1.86 Group 2 (x2 values): n2 = 14 (no caffeine) x2  6.59 and s2  1.91

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

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) + 1.24 11.82 < 1 – 2 < 14.30

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 14.3 hertz is one that contains the difference between caffeine and no caffeine.

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.

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

Example – Solution

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.

Example 10(b) – Solution

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

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.

9.4 Estimating 1 – 2 and p1 – p2 Summarize Notes Read section 9.4 Homework Worksheet