Ch-14-20-HT-CI-Mix - Montgomery College

Math 116 - Chapters 14, 17, 18, 19, 20
Name___________________________________
You may assume that the variable under consideration is normally distributed on both populations. The two population
standard deviations cannot be assumed to be equal (use a NON-POOLED test)
1) A paint manufacturer wishes to compare the drying times of two different types of paint. Independent random
samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying
times were recorded. The summary statistics are as follows.
Type A
Type B
x 1 = 76.0 x 2 = 64.0
s = 4.5
s = 5.1
1
2
n = 11
n =9
1
2
a) Do the data suggest that Type A paint take longer to dry? Test the claim at the 0.01 level of significance.
b) Determine a 98% confidence interval for the difference, µ - µ , between the mean drying time for paint of
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2
types A and B.
c) What is the interval suggesting about µ and µ ? Are they equal or is one of them larger than the other?
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2
A) ANSWER:
B) a) t = 5.517, p = 0.00002 < any alpha. Very small p-value gives a strong evidence against the null
hypothesis and in favor of the alternative hypothesis.
If the mean drying times of the populations are equal (u1 = u2) the chances of of observing such a
difference between the sample means ([(x-bar1) - (x-bar2)] = 12) is almost zero (p = 0.00002). Such a
difference between sample means could be observed easily if mu1 is higher than mu2.
The sample data provides strong evidence to support the claim that Type A paint takes longer to dry.
Notice that this conclusion does not depend of the significance level. At any of the common significance
levels, the conclusion will be the same.
b) We are 98% confident that the difference between the mean drying times for the two groups is
somewhere between 6.38 hs and 17.62 hours.
c) The interval is completely on the positive side, this means µ is larger than µ , which is the same
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conclusion as in the hypothesis testing
2) In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of 220 men, 60%
favored stricter gun control laws.
a) Test the claim that the proportion of women favoring stricter gun control is higher than the proportion of
men favoring stricter gun control. Use a significance level of 0.05.
b) Construct a 90% confidence interval estimate for p1 - p2.
c) What is the interval suggesting?
A) ANSWER:
a) Test statistic: z = 1.21.
p-value = .113. It is very likely (p = 0.113 > any alpha) to observe such a difference between the p-hats (.65
- .60 = .05) when the population proportions are equal.
There is not sufficient evidence to support the claim that the proportion of women favoring stricter gun
control is higher than the proportion of men favoring stricter gun control.
b) We are 90% confident that the difference between the proportions of women and men favoring stricter
gun control legislation is in the interval: (-.0183, .11827)
c) The interval provides plausible values for p1 - p2. Since the interval contains zero, it is possible for p1
to be equal to p2. Notice that it agrees with the conclusion of the hypothesis testing.
B)
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You may assume that variable under consideration is normally distributed on both populations. The two population
standard deviations cannot be assumed to be equal. (use a NON-POOLED test)
3) A researcher was interested in comparing the amount of time spent watching television by women and by men.
Independent random samples of 14 women and 17 men were selected and each person was asked how many
hours he or she had watched television during the previous week. The summary statistics are as follows.
Sample 1 (women) Sample 2 (men)
x 1 = 12.6
s = 3.9
1
n = 14
1
x 2 = 13.6
s = 5.2
2
n = 17
2
a) Do the data provide sufficient evidence to conclude that on average, women watch less hours of TV per week
than men? Test at the 5% significance level.
b) Construct a 90% confidence interval estimate for the difference µ - µ
1
2
c) What is the interval suggesting about µ and µ ?
1
2
A) a) Test statistic: t = -0.611
p-value = 0.273. It is very likely (p = 0.273 > any alpha) to observe such a difference between the sample
means [(x-bar1) - (x-bar2) = -1] when the means of the populations are equal.
At the 5% significance level, the data do not provide sufficient evidence to conclude that on average,
women watch less hours of TV per week than men.
b) We are 90% confident that the difference between the mean times of of the two populations is in the
interval (-3.781 hours , 1.7807 hours)
c) The interval provides plausible values for u1 - u2 and since the interval contains zero, it's possible for
the two means to be equal. Same conclusion as stated in part (a)
B)
Identify the distribution of the sample mean. In particular, state whether the distribution of x is normal or approximately
normal and give its mean and standard deviation.
4) The heights of people in a certain population are normally distributed with a mean of 65 inches and a standard
deviation of 3.4 inches. Determine the sampling distribution of the mean for samples of size 42.
A) Normal, mean = 65 inches, standard deviation = 0.52 inches
B)
C)
D)
5) You wish to estimate the proportion of all voters in California who plan to vote in favor of a certain ballot
measure. How large of a sample should we select to be 99% confident that the sample proportion x-bar is
within 0.015 from the population proportion p.
a) Assume that in a preliminary study 29% of the voters sampled voted in favor.
b) Assume we have no results from a preliminary study.
A) a) 6068, b) 7368; If there is an estimate from a previous study we prefer to use it because it produces a
smaller sample size, which implies a cheaper study.
B)
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6) A manufacturer considers his production process to be out of control when defects exceed 3%. In a random
sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and
production is not really out of control.
a) At the 0.05 level of significance, do the data provide sufficient evidence that the percentage of defects exceeds
3%?
b) Construct a 90% confidence interval estimate for the proportion of defective parts.
c) What does the interval suggest?
A) a) Test statistic: z = 1.57.
p = .0596. If the percent defective is 3%, at the 5% significance level it is likely to observe a p-hat of 5.9% or
higher (p = 0.0596 > 0.05)
At the 5% significance level, there is not sufficient evidence to conclude that the percentage of defects
exceeds 3%.
b) (.0168, .1008). We are 90% confident that the proportion of defects is between 0.0168 and 0.1008.
c) Since the interval contains 0.03, it is possible for the proportion of deffective to be 0.03. Same conclusion
as in part (a)
B)
7) The forced vital capacity (FVC) is often used by physicians to assess a person's ability to move air in and out of
their lungs. It is the maximum amount of air that can be exhaled after a deep breath. For adult males, the
average FVC is 5.0 liters. A researcher wants to perform a hypothesis test to determine whether the average
forced vital capacity for women differs from this value. The mean forced vital capacity for a random sample of
85 women was 4.8 liters.
a) Do the data provide sufficient evidence to conclude that the mean forced vital capacity for women differs
from the mean value for men of 5.0 liters? Perform the appropriate hypothesis test using a significance level of
0.05. Assume that = 0.9 liters.
b) Construct a 95% confidence interval estimate for the mean forced vital capacity of women.
c) What does the interval suggest?
A) a) z = -2.05
P-value = 0.0404. It is unlikely (p = 0.04 < 0.05) to observe a sample mean of 4.8 or a more extreme one
when the population mean is 5.0.
At the 5% significance level, the data provide sufficient evidence to conclude that the mean forced vital
capacity for women differs from 5.0 liters.
b) (4.6087, 4.9913). We are 95% confident that the mean forced vital capacity for women is between 4.6087
and 4.9913.
c) The interval is completely below 5. With 95% confidence we can say that the mean forced vital capacity
for women differs from 5.0 liters.
B)
8) A survey of 865 voters in one state reveals that 459 favor the death penalty.
a) At the 1% significance level can we support the claim that the majority of voters in the state favor the death
penalty?
b) Construct the 98% confidence interval for the true proportion of all voters in the state who favor the death
penalty. Does the interval support the claim stated in (a)? Explain.
A) a) z = -1.8
p = 0.036
Do not reject Ho. At the 1% significance level we do not have enough evidence to support the claim that.....
b) from 0.49116 to .57011
Since the interval contains .50, it may be equal to it. Same conclusion as in part (a).
B)
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Assume that the differences are normally distributed.
9) A coach uses a new technique in training middle distance runners. The times for 8 different athletes to run 800
meters before and after this training are shown below.
Athlete
A
B
C
D
E
F
G
H
Time before
training (seconds) 116.9 108.4 114.1 112.8 116.9 117.1 114.8 111.5
Time after training
(seconds) 117.5 107.1 111.7 113.6 115.1 117.2 111.2 107.6
a) Do the data suggest that the training helps to improve the athletes' times for the 800 meters? Perform a paired
t-test at the 5% significance level.
b) Construce a 90% confidence interval estimate for the mean of the differences.
c) What does the interval suggest?
A) You need to create a List 3 = Time before - time after and perform a T-Test into L3
u(d) = 0 versus u(d) > 0. The sample mean difference d-bar = 1.4375
a) Test statistic: t = 2.227
p = 0.03. It is unlikely (p = 0.03 < 0.05) to observe a sample mean difference d-bar = 1.4375 when mu(d) is
zero, that is, when there is no difference in times before and after the training. Since d-bar is significantly
larger than zero at the 5% significance level, the data provide sufficient evidence to conclude that the
training helps to improve the athletes' times for the 800 meters.
b) We are 90% confident that the mean difference mu(d) is somewhere between .21441 seconds and 2.6606
seconds
c) Since all values in the interval are possitive, the interval suggests that the mean difference is greater
than zero which implies that the training improved the athletes times. Same conclusion as in (a)
B)
10) A light-bulb manufacturer advertises that the average life for its light bulbs is 900 hours. A random sample of
15 of its light bulbs resulted in the following lives in hours.
995
590
510
539
739
917
571
555
916
728
664
693
708
887
849
a) Construct a box plot and a normal probability plot to confirm that there are no outliers and that the
population is approximatelly normally distributed.
b) At the 10% significance level, do the data provide evidence that the mean life for the company's light bulbs
differs from the advertised mean?
c) Construct a 90% confidence interval estimate for the life of the light bulbs. What does the interval suggest?
A) a) Test statistic: t = -1.87.
p = 0.0409 < 0.10. It is unlikely (p = 0.0409 < 0.10) to observe an x-bar of .............. if the mean of the
population is 900 hours. There is sufficient evidence to support the claim that the true mean life differs
from the advertised mean. (Notice that at the 1% significance level the conslusion is different)
b) We are 90% confident that the average life for these light bulbs is somewhere between 652.7 hours and
795.43 hours.
c) Since the interval is completely below 900, it suggests that the mean life time is actually lower than 900
hours, which supports the claim that it is different from 900 hours.
B)
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11) Ten different families are tested for the number of gallons of water a day they use before and after viewing a
conservation video. Do the data suggest that the mean amount after the viewing is different from the mean
amount before the viewing?
a) Perform a paired t-test at the 5% significance level.
b) Construce a 90% confidence interval estimate for the number of gallons of water used per day.
c) What does the interval suggest?
Before 33 33 38 33 35 35 40 40 40 31
After 34 28 25 28 35 33 31 28 35 33
A) a) Test statistic: t = 2.894
p = 0.018
d-bar = 4.8
At the 5% significance level, the data provide sufficient evidence to conclude that the mean amount after
the viewing differs from the mean amount before the viewing.
b) We are 90% confident that .................................... 1.7595 to 7.8405
c) The interval is completely above zero. It suggests that the mean difference is larger than zero which
supports the claim that after viewing the video the families consume a different amount of water a day.
B)
12) A newspaper in a large midwestern city reported that the National Association of Realtors said that the mean
home price last year was $116,800. The city housing department feels that this figure is too low. They randomly
selected 66 home sales and obtained a sample mean price of $118,900. Assume that the population standard
deviation is $3,700.
a) Using a 5% level of significance, perform a hypothesis test to determine whether the population mean is
higher than $116,800.
b) Construct a 90% confidence interval estimate for the mean home price in the midwestern city.
c) What is the confidence interval suggesting about the mean home price of houses in the area?
A) a) Test statistic: z = 4.61.
p = 0.000002. It is very unlikely (p = 0.000002 to observe a sample mean of 118,900 or higher when the
population mean is 116,800. This sample mean gives a strong evidencethat the mean home price is now
higher than than $116,800.
b) We are 90% confident that the mean home price in this large midwestern city is between $118151 and
$119649.
c) Since the interval is completely above 116,800, it suggests that the mean home price is now higher than
$116,800. Same conclussion as in part (a)
B)
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