Review Chapter 8 : CI and Sig. Tests for p and p1-p2 using the z
1. Why do we use the z table?
2. What assumptions do we need to make?
· Random Samples
· Check specific assumption for that test
OR Use the most restrictive one for all cases and you will be ok.
§ One Sample Proportion CI -sample size must be at least equal to 5.
§
§ Two Sample Proportion CI - both sample sizes must be at least equal to 10.
§ Two Sample Proportions ST – at least 5 successes and 5 failures must in each of the samples.
3. Which estimator of p do we use?
·
One Sample Proportion (Confidence
Interval) 
·
One Sample Proportion (Significance
Test) 
- Two
Sample Proportion (Confidence Interval)
- Two
Sample Proportions (Significance Test)
- Make and interpret CI and
ST.
- Read Minitab Output
- Interpret Results
Examples: For each of the following
examples, we will first determine which kind of problem it is. Then we
will set the problem up as if to do by hand.
We will interpret the Minitab output given for each problem.
1. Suppose that in a random sample of 50 babies conceived by an in vitro fertilization process (test tube babies), 35 are girls. Does this sample evidence show convincingly that the in vitro process favors the female gender? Explain.
· Which kind of problem is this?
·
Interpret the results of the Minitab output from
the last page.
2. Four sets of identical twins were selected at random. One child was selected at random from each pair to form an "experimental group"- these four children were sent to school. The other four children were kept at home as a control group. At the end of the school year the IQ scores presented below were obtained. Does this evidence justify the conclusion that lack of school experience has a depressing effect on IQ scores?
Experimental
Control
Pair
Group
Group
A
110 111
B
125
120
C
139
128
D
142
135
· Which kind of problem is this?
·
Interpret the results of the Minitab output from
the last page.
3. A restaurant has recently started using a dessert
cart that is pushed around the eating area for diners to view and, it is hoped,
be unable to resist. On the first ten evenings the cart
was used, the average expenditure per night for
desserts was $130 compared to an average of $110 prior to the use of the
cart. The standard deviation for the ten evenings was $65. Does it
look like the cart will improve sales significantly?
· Which kind of problem is this?
·
Interpret the results of the Minitab output from
the last page.
4. A social scientist believes that the fraction of
Republicans in favor of the death penalty is greater than the fraction of
Democrats in favor of the death penalty. She asked random samples of 200
Republicans and 200 Democrats and found 46 Republicans and 34 Democrats favoring
the death penalty. Do these data support the social scientist's belief?
· Which kind of problem is this?
·
Interpret the results of the Minitab output from
the last page.
5. The effect of alcohol consumption on the body appears to be much greater at high altitudes than at sea level. To test this theory, a scientist randomly selects 12 subjects and randomly divides them into two groups of six each. One group is transported to an altitude of 12,000 feet, where each subject ingests a drink containing 100 cc of alcohol. The second group receives the same drink at sea level. After two hours, the amount of alcohol in the blood (grams per 100 cc) for each subject is measured. The data are shown in the table. Do the data provide sufficient evidence to support the theory that retention of alcohol in the blood is greater at high altitudes?
|
Sea Level |
12,000 Feet |
|
0.07 |
0.13 |
|
0.10 |
0.17 |
|
0.09 |
0.15 |
|
0.12 |
0.14 |
|
0.09 |
0.10 |
|
0.13 |
0.14 |
· Which kind of problem is this?
·
Interpret the results of the Minitab output from
the last page.
Test
and Confidence Interval for One Proportion
Test of p = 0.5 vs
p > 0.5
Exact
X N Sample p 95.0 %
CI P-Value
35 50 0.700000 (0.553918, 0.821382) 0.003
Test
and CI for One Proportion
Sample X
N Sample
p 95.0% CI
1 37
54 0.685185 (0.544476, 0.804761)
Paired T-Test and Confidence Interval
Paired T for exp - control
N Mean StDev
SE Mean
exp 4
129.00 14.67 7.34
control 4
123.50 10.34 5.17
Difference 4
5.50 5.00 2.50
95% CI for mean difference: (-2.46, 13.46)
T-Test of mean difference = 0 (vs
>0): T-Value = 2.20 P-Value = 0.058
T-Test of the Mean
Test of mu =
110.0 vs mu > 110.0
Variable N
Mean StDev SE
Mean T P
dessert 10 130.0 65.3
20.6 0.97 0.18
T
Confidence Intervals
Variable N Mean StDev SE Mean 95.0 % CI
dessert 10 130.0 65.3 20.6
(83.3,176.7)
Test
and Confidence Interval for Two Proportions
Sample
X N Sample p
1 46 200
0.230000
2 34 200 0.170000
Estimate for p(1) - p(2):
0.06
95% CI for p(1) - p(2): (-0.0181778, 0.138178)
Test for p(1) - p(2) = 0 (vs > 0): Z = 1.50
P-Value = 0.066
Test
and CI for Two Proportions
Sample X
N Sample
p
1 47 202 0.232673
2 35
202 0.173267
95%
CI for p(1) - p(2):
(-0.0188205, 0.137632)
Two
Sample T-Test and Confidence Interval
Two sample T for sea level vs 12000ft
N Mean StDev
SE Mean
sea leve 6 0.1000
0.0219 0.0089
12000ft 6 0.1383 0.0232 0.0095
95% CI for mu sea leve - mu 12000ft:(
-0.0678, -0.0089)
T-Test mu sea leve = mu 12000ft (vs <): T = -2.94 P = 0.0082 DF = 9