Download Sample Homework 4

STAT 5352
Probability and Statistics II
Spring 2012
Homework #4
Due on Thursday, March 29, 2012
Instructor: Larry Ammann
CourseBook: John E. Freund’s Mathematical Statistics with Applications (7th Edition)
Emrah Cem(exc103320)
1
Emrah Cem
STAT 5352 (Larry Ammann): Homework #4
Contents
Exercise 1
1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
3
Exercise 2
1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
4
4
Exercise 3
4
Exercise 4
4
Exercise 5
5
Exercise 6
1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
5
Exercise 7
5
Exercise 8
5
Exercise 9
1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
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Emrah Cem
STAT 5352 (Larry Ammann): Homework #4
Exercise 1
Experience has shown that the mean number of sick leave days taken per year by employees in a large
corporation is 10 with a standard deviation of 4. Suppose that a random sample of 62 employees who smoke
is selected, and suppose that their sick leave records for the last year show that the average number of sick
leave days taken is 15 with a s.d. of 12.
1
Answer:
given by,
An interval estimation for population average µ with confidence level 1-α is
√
X̄ ± t(S/ n)
, so 95% confidence interval for the mean number of sick leave days for all employees who
smoke is
12
15 ± t0.025,61 √ = [15 − 2 ∗ 1.524, 15 + 2 ∗ 1.524] = [11.951, 18.048]
62
Exercise 1
(1)
2
Answer: We have n=62 and s2 = 144, so 95% confidence interval for the standard
deviation of the number of sick leave days for all employees who smoke is:
#
"s
#
"r
s
r
61 ∗ 144
61 ∗ 144
61 ∗ 144
61 ∗ 144
,
(2)
,
=
χ20.025,61
χ20.975,61
84.476
41.3
h√
i
√
=
103.98, 212.68
(3)
=
[10.19, 14.58]
Exercise 1
(4)
3
Answer: 95% of all samples of 62 employees who smoke has a higher average number of
sick leave days when compared to the overall average. In other words, randomly selected
62 employees will have a higher average number of sick leave days when compared to the
overall average with probability 0.95.
Exercise 1
Exercise 2
1
Answer: 90% confidence interval for the mean weight loss after 1 year is:
8.2
5.4 ± z0.05 √ = [5.4 − 1.645 ∗ 1.64, 5.4 + 1.645 ∗ 1.64] = [2.7022, 8.097]
25
(5)
we can conclude that average loss weight of randomly selected 25 individuals will bew
within [2.7022, 8.097] with probability 0.95. Note that this does not mean anyone who
joined this weight-loss center will lose weight since the range does not contain negative
values. Randomly selected 25 people may have negative average weight loss.
Exercise 2 continued on next page. . .
Exercise 2
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Emrah Cem
STAT 5352 (Larry Ammann): Homework #4
Exercise 2 (continued)
2
Answer: 90% confidence interval for the standard deviation of weight loss after 1 year
is:
"s
#
"r
#
s
r
24 ∗ 67.24
24 ∗ 67.24
24 ∗ 67.24
24 ∗ 67.24
,
=
(6)
,
χ20.05,24
χ20.95,24
36.415
13.848
=
[6.657, 10.795]
Exercise 2
(7)
3
Answer: sample size should be
1.96∗8.2 2
1
≈ 258
Exercise 2
Exercise 3
A large corporation requires that its employees attend a 1-day sexual harassment seminar. The Director of
Human Resources of this corporation would like to determine whether or not the information presented in
this seminar is retained over a long period of time. To this end, a random sample of 30 employees is randomly
selected from recently hired employees who are scheduled to take this seminar. Each of the employees in
this sample completes a test of knowledge concerning sexual harassment and related legal issues immediately
after the seminar, and then takes a similar test 6 months later. The scores are summarized below. Construct
a 95% confidence interval for the difference between the mean scores. Mean(first test) = 83.6, s.d.(first test)
= 12.4 mean(second test) = 74.8, s.d.(second test) = 15.8 s.d.(differences) = 18.1.
Answer: We are given
X̄1 − X̄2 = 8.8
sX̄1 −X̄2 = 18.1
s1 = 12.4
s2 = 15.8
Lets first calculate the degrees of freedom. The following formula is appropriate whenever a t score is used
to analyze the difference between means. DF = (s21 /n1 + s22 /n2 )2 /[(s21 /n1 )2 /(n1 − 1)] + [(s22 /n2 )2 /(n2 − 1)]
In our case this value is approximately 55.
95% CI = 8.8 ±t0.975,55 18.1 = [8.8 − 36.2 + 8.8 + 36.2] = [−27.4, 45]
Exercise 4
A corporation would like to convert its training courses from instructor-led classroom courses to individual
computer instruction to reduce costs. To determine if there is a difference in performance of these two training
methods, a sample of 30 new employees was randomly assigned to two groups of 15 each. The first group
received the standard classroom training and the second group received the individual computer instruction.
After completing the training, all students took the post-training exam with the following results: X̄1 = 84.4
, s1 = 12.5 , X̄2 = 79.1 , s2 = 10.6 . Construct a 90% confidence interval for the difference between the
mean scores of the two groups. You may assume that the histograms of the test scores are approximately
normal curves.
Answer: 90% CI =
r
12.52 + 10.62
(84.4 − 79.1) ± 1.64 ∗
= [5.3 − 6.94, 5.3 + 6.94] = [−1.64, 12.24]
15
Exercise 4 continued on next page. . .
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Emrah Cem
STAT 5352 (Larry Ammann): Homework #4
Exercise 4 (continued)
Exercise 5
A researcher wishes to estimate with 90% confidence the proportion of employees in a particular industry
who favor random drug testing of all employees in this industry. She also would like to have her estimate
be within 0.02 of the true value with this level of confidence. What sample size should she use if she is not
willing to place any prior bounds on this proportion? What sample size should she use if she believes that
this proportion will be at least 75%?
1.645 2
1.15 2
Answer: For 90% CI, n should be 2∗0.02
≈ 1691, For 75% CI, n should be 2∗0.02
≈ 826
Exercise 6
Suppose that the mean IQ scores of 12 year olds with a particular learning disability is 72, and you wish to
determine if a new protocol for educating these students can improve their IQ. A random sample of 24 such
children receive this new protocol, and are then tested 1 month after completion. Suppose that these results
show a mean IQ for these 24 students is 75 with a standard deviation of 12.
1
Answer: 95% CI for the mean IQ of students who receive this protocol is 75±1.96∗12 = [51.48, 98.52]
2
Answer: n =
zα/2×σ 2
e
=
1.96×12 2
3
≈ 62
Exercise 7
A survey of 600 owners of 2010 Ford Explorers showed that 375 of these owners were satisfied with the reliability of this vehicle. Construct a 90% confidence interval for the proportion of all such owners who are satisfied
with the reliability of this car model. What sample size would be necessary to estimate this proportion to
within .02 with 95% confidence if no assumptions are made regarding this population proportion?
Answer: 90% CI for
q the proportion of all such owners who are satisfied with the reliability of this car
model is :
375
600
±1.645
375 225
600 600
= 0.625±0.032 = [0.593, 0.657]. In order to estimate this proportion to within
2
.02 with 95% confidence, sample size should be 1.96
0.625 ∗ 0.375 ≈ 2250. If the sample standart deviation
.02
2
is not representative for population standart deviation then sample size should be 1.96
0.5 ∗ 0.5 ≈ 2401.
.02
600
Exercise 8
Suppose that the mean time to complete a certain task has approximately a normal distribution with mean
of 8.3 minutes and s.d. 1.8. After rearranging the order of some of the tasks that make up the job, it was
found that the mean completion time of a random sample of 18 such jobs was 7.6 minutes with a s.d. of
1.2 minutes. You may assume that these times are approximately normally distributed. Construct a 90%
Exercise 8 continued on next page. . .
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Emrah Cem
STAT 5352 (Larry Ammann): Homework #4
Exercise 8 (continued)
confidence interval for the mean completion time after the rearrangement and for the s.d. of the completion
time. What conclusions can you make?
Answer: 90% confidence interval for the mean completion time after the rearrangement is 7.6±1.645∗ √1.2
=
18
7.6 ± 0.465 = [7.135, 8.065]
q
q
2
2
90% confidence interval for the s.d. of the completion time after the rearrangement is [ χ17∗1.2
, χ17∗1.2
]=
0 .95,17
0 .05,17
√
√
[ 0.887, 2.823] = [0.941, 1.68]
We are 95 % confident that mean time to complete a task has decreased, as well as the s.d of the completion
time. We can also say that, 95% of all samples with size 18 has mean completion time less than 8.3 minutes
and s.d of completion time less than 1.8 minutes.
Exercise 9
A large accounting firm would like to estimate the proportion of errors made by its employees on tax filings.
It randomly selects 250 files and after close examination finds that 20 contain errors.
1
Answer: 90% CI for the overall proportion of errors made on its tax filings is
0.08 ± 0.028 = [0.052, 0.108].
20
250
± 1.645
q
20 230
250 250
250
=
2
Answer: 90 % CI for the mean dollar value is 425 ± 1.645 ∗
√80
20
= [395.57, 454.42]
3
Answer: 90% confidence interval for the s.d. is [
q
19∗802
χ0 .95,19 ,
q
19∗802
χ0 .05,19 ]
= [63.514, 109.632]
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