Question 1:
Consider the following set of data.
|
Pairs
|
1
|
2
|
3
|
4
|
5
|
|
Sample A
|
9
|
4
|
3
|
5
|
3
|
|
Sample B
|
3
|
8
|
2
|
7
|
1
|
(a) Find the paired differences, d = A - B, for this set of data.
(d1)
(d2)
(d3)
(d4)
(d5)
(b) Find the mean d of the paired differences. (Give your answer correct to one decimal place.)
(c) Find the standard deviation sd of the paired differences. (Give your answer correct to two decimal places.)
Question 2:
Salt-free diets are often prescribed to people with high blood pressure. The following data values were obtained from an experiment designed to estimate the reduction in diastolic blood pressure as a result of consuming a salt-free diet for 2 weeks. Eight subjects had their blood pressure measured and then ate a salt free diet for two weeks and had their blood pressure measured again. Assume diastolic readings to be normally distributed.
|
Before
|
99
|
105
|
93
|
102
|
100
|
108
|
107
|
97
|
|
After
|
92
|
102
|
91
|
94
|
96
|
98
|
100
|
93
|
(a) The proper TI-83 program to use to compute the confidence interval for the mean reduction in blood pressure is:
(b) Find the 98% confidence interval for the mean reduction. (Give your answers correct to two decimal places.)
(c) Which of the following statements is true about the confidence interval? (More than one may apply)
Question 3:
An experiment was designed to estimate the mean difference in weight gain for pigs fed ration A as compared with those fed ration B. Eight pairs of pigs were used. The pigs within each pair were litter-mates. The rations were assigned at random to the two animals within each pair. The gains (in pounds) after 45 days are shown in the following table. Assuming weight gain is normal, find the 99% confidence interval estimate for the mean of the differences μd, where d = ration A - ration B. (Give your answers correct to two decimal places.)
|
Litter
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
|
Ration A
|
56
|
40
|
60
|
59
|
43
|
40
|
50
|
46
|
|
Ration B
|
54
|
30
|
50
|
56
|
37
|
36
|
42
|
40
|
| Litter |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
Total |
mean |
| Ration A |
56 |
40 |
60 |
59 |
43 |
40 |
50 |
46 |
394 |
49.25 |
| Ration B |
54 |
30 |
50 |
56 |
37 |
36 |
42 |
40 |
345 |
43.125 |
| d |
2 |
10 |
10 |
3 |
6 |
4 |
8 |
6 |
49 |
6.125 |
| dsquared |
4 |
100 |
100 |
9 |
36 |
16 |
64 |
36 |
365 |
mean sum of square |
|
|
|
|
|
|
|
|
|
|
52.14286 |
(a) There is an increase in the mean difference between post-test and pretest scores.(d=post-test scores - pretest scores)
(b) Following a special training session, it is believed that the mean of the difference in performance scores will not be zero.
(c) On average, there is no difference between the readings from two inspectors on each of the selected parts.
(d) The mean of the differences between pre-self-esteem scores and post-self-esteem scores showed improvement after involvement in a college learning community. (d= post self-esteem scores - pre-self-esteem scores.)
Question 4:
Ten randomly selected college students, who participated in a learning community, were given pre-self-esteem and post-self-esteem surveys. A learning community is a group of students who take two or more courses together. Typically, each learning community has a theme, and the faculty involved coordinate assignments linking the courses. Research has shown that the benefits of higher self-esteem, higher grade point averages (GPAs), and improved satisfaction in courses, as well as better retention rates, result from involvement in a learning community. The scores on the surveys are as follows.
|
Student
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
|
Prescore
|
21
|
19
|
14
|
19
|
18
|
16
|
11
|
20
|
20
|
16
|
|
Postscore
|
16
|
13
|
12
|
22
|
19
|
14
|
16
|
13
|
15
|
16
|
|
Student
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Total
|
means
|
|
Prescore
|
21
|
19
|
14
|
19
|
18
|
16
|
11
|
20
|
20
|
16
|
174
|
17.4
|
|
Postscore
|
16
|
13
|
12
|
22
|
19
|
14
|
16
|
13
|
15
|
16
|
156
|
15.6
|
|
d
|
5
|
6
|
2
|
-3
|
-1
|
2
|
-5
|
7
|
5
|
0
|
18
|
1.8
|
|
dsquared
|
25
|
36
|
4
|
9
|
1
|
4
|
25
|
49
|
25
|
0
|
178
|
19.77778
|
(a) The proper TI-83 Program to solve this problem is:
(b) Find the p-value. (Give your answer correct to three decimal places.)
(c) State the appropriate conclusion.
Question 5:
To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in 1 minute, both before and after the course, was recorded. Ten randomly selected participants scored as shown in the following table. Can you conclude that a significant amount of improvement took place? )(d=after - before) Use α = 0.01 and assume normality.
|
Before
|
11
|
19
|
16
|
11
|
11
|
13
|
19
|
13
|
15
|
17
|
|
After
|
21
|
25
|
23
|
23
|
24
|
25
|
25
|
19
|
15
|
21
|
|
|
|
|
|
|
|
|
|
|
|
total
|
mean
|
|
Before
|
11
|
19
|
16
|
11
|
11
|
13
|
19
|
13
|
15
|
17
|
145
|
14.5
|
|
After
|
21
|
25
|
23
|
23
|
24
|
25
|
25
|
19
|
15
|
21
|
221
|
22.1
|
|
d
|
10
|
6
|
7
|
12
|
13
|
12
|
6
|
6
|
0
|
4
|
76
|
7.6
|
|
dsquared
|
100
|
36
|
49
|
144
|
169
|
144
|
36
|
36
|
0
|
16
|
730
|
81.11111
|
(a) Find the p-value. (Give your answer correct to three decimal places.)
(b) State the appropriate conclusion.
(c) Which of the following statements are true about the p-value?
1. P-value is the probability that our statistical evidence, the sample mean is explained by sampling variation or "chance"
2. The p-value is the probability that the null hypothesis is true
3. The p-value is a probability statement about the sample mean
4. p-value is calculated using a t-sampling distribution with 9 df with μd = 0
5. The p-value is the probability that the null hypothesis is false
Question 6:
Two men, A and B, who usually commute to work together, decide to conduct an experiment to see whether one route is faster than the other. The men believe that their driving habits are approximately the same, and therefore they decide on the following procedure. Each morning for 2 weeks, A will drive to work on one route and B will use the other route. On the first morning, A will toss a coin. If heads appear, he will use route I; if tails appear, he will use route II. On the second morning, B will toss the coin: heads, route I; tails, route II. The times, recorded to the nearest minute, are shown in the following table. Assuming commute times are normal, find the 95% confidence interval estimate for the mean of the differences μd, where d = route I - route II. (Round your answers to two decimal places.)
|
Day
|
|
Route
|
M
|
Tu
|
W
|
Th
|
F
|
M
|
Tu
|
W
|
Th
|
F
|
|
I
|
29
|
29
|
25
|
30
|
27
|
24
|
25
|
29
|
24
|
29
|
|
II
|
29
|
26
|
28
|
26
|
25
|
23
|
23
|
29
|
30
|
25
|
Question 7:
Reduced heart rate variability (HRV) is known to be a predictor of mortality after a heart attack. One measure of HRV is the average of normal-to-normal beat interval (in milliseconds) for a 24-hour time period. Twenty-two heart attack patients who were dog owners and 50 heart attack patients who did not own a dog participated in a study of the effect of pet ownership on HRV, resulting in the summary statistics shown in the accompanying table.
|
Measure of
HRV (average
normal-to-normal
beat interval)
|
|
Mean
|
Standard
Deviation
|
|
Owns Dog
|
873
|
136
|
|
Does Not Own Dog
|
805
|
133
|
The authors of this paper used a two-sample t test to test H0: μ1 - μ2 = 0 versus Ha: μ1 - μ2 ≠ 0.
(a) What assumptions must be made in order for this to be an appropriate method of analysis?
(b) The paper indicates that the null hypothesis was rejected and reports that the P-value is less than 0.01. Carry out the two-sample t test. What is your p-value rounded to 3 decimal
(c) Is your conclusion consistent with that of the paper?
Yes No
Question 8:
Each person in random samples of 237 male and 263 female working adults living in a certain town in Canada was asked how long, in minutes, his or her typical daily commute was.
|
Males
|
Females
|
|
Sample
size
|
x1
|
s
|
Sample
size
|
x2
|
s
|
|
237
|
31.6
|
24.0
|
263
|
29.3
|
24.3
|
Is there enough evidence to show that there is a difference in mean commute times for male and female working residents of this town? Use a significance level of 0.01.
(a) The appropriate alternate hypothesis is HA:
(b) The appropriate TI-83 program to use for this analysis is because the population variance isn't known
(c) df (rounded to nearest whole number)
(d) p-value (rounded to three decimal places)
(e) The p-value is the probability that the difference between x1 - x2 and is explained by if H0 is
(f) Do you have strong enough evidence to reject H0?
Yes No
Question 9:
The paper investigated the driving behavior of teenagers by observing their vehicles as they left a high school parking lot and then again at a site approximately mile from the school. Assume that it is reasonable to regard the teen drivers in this study as representative of the population of teen drivers. Assume the respective populations are normal.
|
Amount by Which Speed Limit Was Exceeded
|
|
Male
Driver
|
Female
Driver
|
|
1.4
|
-0.1
|
|
1.2
|
0.4
|
|
0.9
|
1.1
|
|
2.1
|
0.7
|
|
0.7
|
1.1
|
|
1.3
|
1.2
|
|
3
|
0.1
|
|
1.3
|
0.9
|
|
0.6
|
0.5
|
|
2.1
|
0.5
|
The measurements represent the difference between the observed vehicle speed and the posted speed limit (in miles per hour) for a sample of male teenage drivers and a sample of female teenage drivers. Use a significance level of 0.01 to test the hypothesis that males exceed the speed limit by more miles per hour than females.
(a) The correct alternative hypothesis is: HA :
(b) df =
(c) P =
(d) Do these data provide convincing support for the claim that, on average, male teenage drivers exceed the speed limit by more than do female teenage drivers?
Yes No
(e) Which of the following statements are true about the p-value.
1. The p-value is a probability
2. The p-value is calculated using a normal sampling distribution
3. P-value is the probability that our statistical evidence, the sample mean difference is explained by sampling variation or "chance"
4. The p-value is the probability that the null hypothesis is false
5. The p-value is a probability statement about the the sample mean of the difference set of two dependent samples
6. The p-value is the probability that the null hypothesis is true and the sample evidence is due to sampling variation
7. The p-value is a probability about the the difference between two sample means
Question 10:
In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work. The samples were selected at random from working and nonworking students at a university. Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed? Use a significance level of 0.01.
|
|
Sample
Size
|
Mean
GPA
|
Standard
Deviation
|
|
Students Who
Are Employed
|
170
|
3.12
|
.485
|
|
Students Who
Are Not Employed
|
116
|
3.23
|
.524
|
Are Not Employed 116 3.23 .524
(a) The correct alternate hypothesis is: HA:
(b) df =
(c) P =
(d) Is there enough evidence to reject the null hypothesis?
Yes No
Question 11:
Many people take ginkgo supplements advertised to improve memory. Are these over-the-counter supplements effective? In a study devoted to this problem, elderly adults were assigned at random to either a treatment group or a control group. The 102participants who were assigned to the treatment group took 40 mg of ginkgo three times a day for six weeks. The 117 participants assigned to the control group took a placebo pill three times a day for six weeks. At the end of six weeks, the Wechsler Memory Scale (a test of short-term memory) was administered. Higher scores indicate better memory function. Summary values are given in the table below.
|
|
n
|
x
|
s
|
|
Gingko
|
102
|
6.1
|
.6
|
|
Placebo
|
117
|
6
|
.6
|
Is there evidence that taking 40 mg of ginkgo three times a day is effective in increasing mean performance on the Wechsler Memory Scale? Test the relevant hypotheses using α = 0.01.
(a) The correct alternate hypothesis is: HA:
(b) df =
(Round to Integer)
(c) P =
(Round the answer to three decimal places.)
(d) Is there enough evidence to reject the null hypothesis?
Yes No
Question 12:
Techniques for processing poultry were examined in an article. Whole chickens were chilled 0, 2, 8, or 24 hours before being cooked and canned. To determine whether the chilling time affected the texture of the canned chicken, samples were evaluated by trained tasters. One characteristic of interest was hardness. The summary quantities were obtained. Each mean is based on 36 ratings.
|
Chilling Time
|
|
0 hr
|
2 hr
|
8 hr
|
24 hr
|
|
Mean Hardness
|
7.52
|
6.56
|
5.69
|
5.65
|
|
Standard Deviation
|
.96
|
1.72
|
1.31
|
1.50
|
(a) Do the data prove that there is a difference in mean hardness for chicken chilled 0 hours before cooking and chicken chilled 2 hours before cooking? Do a hypothesis test. Use α = .05.
(b) Do the data prove that there is a difference in mean hardness for chicken chilled 8 hours before cooking and chicken chilled 24 hours before cooking? Do a hypothesis test. Use α = .05.
yes no
(c) Use a 90% confidence interval to estimate the difference in mean hardness for chicken chilled 2 hours before cooking and chicken chilled 8 hours before cooking. (Round the answers to three decimal places.)
Question 13:
In a certain research study, 11 male meadow voles who had a single gene introduced into a specific part of the brain were compared to 20 male meadow voles who did not undergo this genetic manipulation. All of the voles were paired with a receptive female partner for 24 hours. At the end of the 24 hour period, the male was placed in a situation where he could choose either the partner from the previous 24 hours or a different female. The percentage of the time during a three-hour trial that the male spent with his previous partner was recorded. The accompanying data are the approximate values read from a graph that appeared in the corresponding article.
Do these data support the researchers' hypothesis that the mean percentage of the time spent with the previous partner is greater for genetically altered voles than for voles that did not have the gene introduced? Test the relevant hypotheses using α = 0.01.
|
|
Percent of Time Spent with
Previous Partner
|
|
Genetically
Altered
|
59, 61, 62, 70, 80, 81, 85, 88,
89, 90, 93
|
|
Not Genetically
Altered
|
6, 15, 17, 21, 22, 24, 38, 43, 53,
61, 64, 71, 75, 78, 82, 85, 87,
94, 98, 100
|
(a) The correct alternate hypothesis is: HA :
(b) df =
(Round to Integer)
(c) P-Value =
(Round the answer to three decimal places.)
(d) Do you have enough evidence to reject the null hypothesis?
Yes No