Download Math 251, Review Questions for Test 3

Math 251, Review Questions for Test 3
Rough Answers
1. (Review of some terminology from Section 7.1) In a state with 459,341 voters, a poll of 2300
voters finds that 45 percent support the Republican candidate, where in reality, unknown to
the pollster, 42 percent support the Republican candidate.
(a) What is the value of the statistic of interest?
Answer. 45% (a statistic is a numerical property of the sample)
(b) What is the value of the parameter of interest?
Answer. 42% (a parameter is a numerical property of the population)
(c) Describe the population of interest.
Answer. The population is all voters in the state.
(d) In general, is it true that given a certain population, the parameter of interest will not
change under repeated sampling? Explain.
Answer. True, the parameter does not depend on a specific sample, so it doesn’t change when
the sample changes.
2. (a) A produce company claims that the mean weight of peaches in a large shipment is 6.0
oz with a standard deviation of 1.0 oz. Assuming this claim is true, what is the probability
that a random sample of 1000 of these peaches would have a mean weight of 5.9 oz or less?
Answer: This is a central limit theorem type problem on sampling distributions (see Section
7.2 for more).
5.9 − 6.0
z=
≈ −3.16
1.0
√
1000
P (z < −3.16) = .5 − .4992 = .0008 Thus, there is approximately a .0008 probability of
obtaining such a sample assuming that the mean is 6.0 oz.
(b) If the store manager randomly selected 1000 peaches and found that the mean weight of
those 1000 peaches was 5.9 oz, should she be suspicious of the produce company’s claim that
the mean weight of peaches in the shipment is 6.0 oz? Explain.
Answer. Yes, she is about 99.9% sure that the true mean weight is less than 6.0 oz.
3. (From p. 349 #6). The heights of 18 year-old men are approximately normally distributed,
with mean 68 inches and standard deviation 3 inches.
1
(a) What is the probability that an 18 year-old man selected at random is between 67 and 69
inches tall.
1
1
Answer. P (67 < x < 69) = P − < z <
= 0.6293 − 0.3707 = 0.2586.
3
3
(b) If a random sample of nine 18 year-old men is selected, what is the probability that the
mean height x̄ is between 67 and 69 inches?
Answer. P (67 < x̄ < 69) = P
−
1
√3
9
<z<
1
!
= P (−1 < z < 1) = 0.6826
√3
9
(c) Why was the probability in (b) higher than that in (a)?
Answer. The sampling distribution standard deviation is smaller than the standard deviation
for the original distribution, so (b) should be higher. In general, one expects that a randomly
chosen group will have an average much closer to the population mean than a randomly
selected individual, the larger the group, the closer one would expect its sample mean to be to
the population mean.
(d) Would you expect the probability that an 18 year-old man selected at random is more than
74 inches tall to be lower, the same as, or higher than the probability of selecting a random
sample of nine eighteen year-old men with a mean height of more than 74 inches?
Answer. Because the sampling distribution has a smaller standard deviation than the original
population distribution, there is a much higher probability of finding individuals far away from
the mean, than random samples whose sample mean is far away from the population mean.
Thus, we expect that it is much more likely to randomly an individual more than 74 inches
tall, than a group of 9 whose mean height is more than 74 inches.
4. (From p. 349 # 18). The taxi and takeoff time for commercial jets is a random variable x
with mean 8.5 minutes and standard deviation of 2.5 minutes. You may assume the jets are
lined up on the runway so that one taxis and takes off immediately after the other, and they
take off one at time on a given runway. What is the probability that for 36 jets on a given
runway total taxi and take off time will be
(a) less than 320 minutes?
Answer. P
320
x̄ <
36
=P
z<
8.39 − 8.5
!
= P (z < .93) = 0.8238.
2.5
√
36
(b) more than 275 minutes?
Answer. P
x̄ >
275
36
=P
z>
7.639 − 8.5
!
2.5
√
36
2
= P (z > −2.07) = 1 − 0.0192 = 0.9808.
(c) between 275 and 320 minutes?
Answer. From solutions (a) and (b), the probability is .8238 − .0192 = .8046.
5. In a Gallup poll it was reported that 47% of adult Americans surveyed approve of the way
the United States has handled the war in Iraq. Moreover, Gallup reported their methods were
as follows:
“These results are based on telephone interviews with a randomly selected national sample of
1,006 adults, aged 18 and older, conducted Oct. 24-26, 2003. For results based on this sample,
one can say with 95% confidence that the maximum error attributable to sampling and other
random effects is 3 percentage points. In addition to sampling error, question wording and
practical difficulties in conducting surveys can introduce error or bias into the findings of public
opinion polls.”
(a) What is the confidence interval that Gallup is suggesting for the proportion of adult Americans favoring the way the U.S. has handled war in Iraq?
Answer. Gallup is claiming that the 95% confidence interval for the proportion is (.44, .50).
(b) Find a 99% confidence interval for the proportion of adult Americans that approve of the
way the United States has handled the war in Iraq.
Answer. We are given p̂ = .47, n = 1006.rNow, q̂ = .53 and clearly np̂ > 5 and nq̂ > 5 so we
pq
. Using the table we find z.99 = 2.58. Therefore,
can use the formulas p̂ ± zc σp̂ where σp̂ =
r n
(.47)(.53)
the 99% confidence interval is .47 ± 2.58
which yields the interval (0.4294, 0.5106).
1006
(c) What sample size would be needed to estimate the proportion of adult Americans that
approve of the way the United States has handled the war in Iraq to within ±.01 with 95%
confidence?
z 2
c
Answer. If we use p = .47 as an estimate for p in the formula n = pq
we get n =
E
2
(.47)(.53) 1.96
= 9569.43 and so we would use n = 9570. If we don’t assume an initial
.01
1 zc 2
estimate for p, we get n =
= 9605.
4 E
6. (a) What sample size would be needed to estimate the mean from a population with
standard deviation of 100 with a maximum error of 5 with 95% confidence?
2
z σ 2
1.96 · 100
c
Answer. We use the formula n =
, and so n =
= 1536.64 and so we
E
5
use n = 1537.
(b) What sample size from the same population would be needed to estimate the mean with a
3
maximum error of 10 with 95% confidence?
2
z σ 2
1.96 · 100
c
Answer. We use the formula n =
, and so n =
= 384.16 (which is
E
10
one-fourth of the answer in (a)). Going to up to the next whole number, we get n = 385.
(c) In general, what effect would increasing a sample size by a factor of 16 have on the maximum
error E?
Answer. Multiplying the sample size by 16 reduces the error to one-fourth of what it was
σ
with the original sample size. This is verified with the following calculation. If E = zc √
n
the error with the original sample size, then the error with a sample 16 times the size of the
original is
1
σ
1 σ
zc √
= zc √ = E.
4
4
n
16n
(d) In general, how much must a sample size be increased to cut the maximum error by
one-half?
Answer.
The
2 sample size must be quadrupled. This is verified with the following calculation.
z σ
c
If n =
is the original sample size, then sample size needed to halve E is
E
!2
z σ 2
zc σ
c
=4
= 4n.
E
E
2
7. In a 1999 survey of 80 Computer Science graduates and 110 Electrical Engineering graduates, it was found that the Computer Science graduates had a mean starting salary of $48, 100
with a standard deviation of $7, 200, while the Electrical Engineering graduates had a mean
starting salary of $52, 900 with a standard deviation of $5, 300.
(a) Find a point estimate for the difference in average starting salaries for Computer Science
and Electrical Engineering graduates.
Answer. The reasonable point estimate is x̄1 − x̄2 = 48, 100 − 52, 900 = −4800.
(b) Let µ1 be the population mean starting salary for the Computer Science graduates and
let µ2 be the population mean salary for the Electrical Engineering graduates. Find a 96%
confidence interval for µ1 − µ2 .
Answer: Since each sample size is at least 30, we use the formula:
s
σ12 σ22
x̄1 − x̄2 ± zc σx̄1 −x̄2 where σx̄1 −x̄2 =
+
n1 n2
4
For this, c = .96 and so z.96 = 2.05 (which is found from the normal table as the z-value for
which P (z < zc ) = .98 since .98 = .5 + .96/2). Now compute
r
7, 2002 5, 3002
σx̄1 −x̄2 =
+
= 950.45
80
110
and so we compute −4800 ± 2.05 · 950.45 which yields the interval (−$6, 748.43, −$2, 851.57).
This means that we are 96% confident that the true difference in salaries is between −$6, 748.43
and −$2, 851.57, i.e. we are very confident that the Computer Science graduates’ starting mean
salary is from $2,851.57 to $6,748.43 less than Electrical Engineering graduates’ mean starting
salaries.
(c) Based on the interval in (b), would you be comfortable saying that the mean starting
salary for Computer Science graduates is less than that for Electrical Engineering graduates?
Explain.
Answer: Yes, we are very confident that it is at least $2,851.57 less than the mean for Electrical
Engineering graduates.
8. (a) A Gallup News Release (see http://www.gallup.com/poll/releases/pr031030.asp) reported that in July 1996, 69% of those surveyed supported assisted suicide for terminally ill,
while in May 2003, 72% of those surveyed supported assisted suicide for the terminally ill.
Assume the 1996 poll surveyed 1103 adult Americans, while the May 2003 poll surveyed 1009
adult Americans. Find a 99% confidence interval for p1 − p2 where p1 is the proportion of
adult Americans supporting assisted suicide in July 1996, and p2 is the proportion of adult
Americans supporting assisted suicide in May 2003.
Answer: First, because n1 p̂1 = (1103)(.69) > 5, n1 q̂1 = (1103)(.31) > 5, n2 p̂2 = (1009)(.72) >
5 and n2 q̂2 = (1009)(.28) > 5, we can use the formula
r
p 1 q1 p 2 q2
+
.
p̂2 − p̂2 ± zc σp̂1 −p̂2 where σp̂1 −p̂2 =
n1
n2
With c = .99 we find z.99 = 2.58 from the table, and we compute
r
r
p̂1 q̂1 p̂2 q̂2
(.69)(.31) (.72)(.28)
σp̂1 −p̂2 =
+
=
+
= .0198426
n1
n2
1103
1009
The 99% confidence interval has endpoints (.69 − .72) ± 2.58 · .0198436, thus the interval is
(−.0812, .0212). We are 99% confident in July 1996 the percentage of adult Americans that
supported assisted suicide was 8.12% less to 2.12% more than the percentage in May 2003.
(b) Based on your interval in (a) would you be comfortable in saying that the proportion
of adult Americans supporting assisted suicide was higher in May 2003 than in July 1996?
Explain.
Answer: No, I could not say that with 99% confidence because the interval found in (a)
included the possibility that the support was up to 2.12% higher in July 1996 than in May
2003.
5
9. (From p. 385 #6) The Roman Arches is an Italian restaurant. The manager wants to
estimate the average amount a customer sends on lunch Monday through Friday. A random
sample of 115 customers’ lunch tabs gave a mean of x̄ = $9.74 with a standard deviation
s = $2.93.
(a) Find a 95% confidence interval for the average amount spent on lunch by all customers.
Answer. The sample size is n = 115 ≥ 30, and so we use the formula for large sample means,
2.93
i.e. x̄ ± zc σx̄ . In this case, σx̄ = √
= .2732 and z.95 = 1.96 and so the 95% confidence
115
interval is ($9.20, $10.28).
(b) For a day when the Roman Arches has 115 lunch customers, use part (a) to estimate the
range of dollar values for the total lunch income that day.
Answer. We are 95% confident that total would be in the range (115 · $9.20, 115 · $10.28),
that is ($1058, $1182).
10. (From p. 398 #12) The number of calories for 3 ounces of french fries at eight popular
fast food chains are as follows.
222 255 254 230 249 222 237 287
Use these data to find a 99% confidence interval for the mean calorie count in 3 ounces of
french fries obtained from fast-food restaurants.
Answer. First, one needs to compute the sample mean and sample standard deviation for the
8 numbers above. For this, one finds
X
x = 1956
X
1956
= 244.5 and s =
Thus x̄ =
8
x2 = 481548
r
SSx = 481548 −
19562
= 3306.
8
3306
= 21.73.
7
This is a small sample (n < 30) with unknown standard deviation, and we’ll assume the
population is normal, or nearly so. We use the t-distribution with 7 degrees of freedom, thus
s
t.99 = 3.499, and the endpoints of the confidence interval are given by x̄ ± tc √ . Thus the 99%
n
21.73
confidence interval has endpoints 244.5 ± 3.499 √ which yields the interval (217.6, 271.4).
8
11. (a) Under are the conditions necessary for finding a confidence interval for the mean from
a small sample?
Answer. We should use the t-distribution with n − 1 degrees of freedom when the sample size,
n, is less than 30 provided the population is normal (or nearly normal) and the population
standard deviation is not known.
6
(b) Under what conditions can you find a confidence interval for the mean using a large sample?
Answer. If the sample size is 30 or larger. In some asymmetric distributions a larger sample
size may be necessary. However, for many practical problems, n ≥ 30 is a good rule of thumb.
(c) What are the conditions necessary for finding a confidence interval for a population proportion?
Answer. We need np > 5 and nq > 5, for this we check that np̂ > 5 and nq̂ > 5, i.e. there
are more than 5 successes and more than 5 failures in the sample. (Notice if p is very close to
0, or very close to 1 the binomial distribution is highly nonsymmetric about its mean, and so
values of n much larger than 30 will be needed to guarantee np > 5 and nq > 5, so do not use
the rule of thumb n ≥ 30 when dealing with binomial distributions.)
12. A recent Gallup poll of 1006 adult Americans found that 37% of those asked oppose
cloning of human organs.
(a) Find a 98% confidence interval for the proportion of adult Americans that oppose cloning
of human organs.
Answer. First z.98 = 2.33 (use table), p̂ = .37 and n = 1006. Now np̂ > 5 and nq̂ > 5 and so
we can use the confidence interval formula for proportions since the sampling distribution will
be approximately normal. Therefore, the 98% confidence interval has endpoints
r
(.37)(.63)
.37 ± 2.33
1006
That gives a confidence interval of (.3345, .4055).
(b) Based on the answer to (a), would it be appropriate for the Gallup organization to state
that less than 38% of adult Americans oppose cloning of human organs? Explain.
Answer. Because the upper end of the confidence interval is 40.55% which is well above 38%,
it would not be appropriate to state that less that 38% of all adult Americans oppose cloning
of human organs using this confidence interval.
13. On June 7, 1999 a poll on the USA Today website showed that out of 2000 respondents,
71% felt that Andre Agassi deserved to be ranked among the greatest tennis players ever.
(a) Assuming that the 2000 respondents form a random sample of the population of tennis
fans, construct a 95% confidence interval for the proportion of all tennis fans who feel that
Andre Agassi should be ranked among the greatest tennis players ever.
Answer. The confidence interval is (.6901, .7299). To find this confidence interval we found
z.95 = 1.96, q
p̂ = .71, q̂ = .29 and n = 2000. Clearly np̂ > 5 and nq̂ > 5 and so we computed:
.71 ± 1.96 ·
(0.71)(0.29)
2000
= .71 ± .0199 to find the endpoints of the interval.
7
(b) Based on (a), would you be comfortable in saying that the poll is accurate to within plus
or minus 2 percent 19 times out of 20? Explain.
Answer. Yes, the 95% confidence interval is .71 ± .0199, hence intervals based on this size
of random sample with the given proportion should have an accuracy of ±2% on average 19
times out of 20.
(c) In actuality, the survey was based on voluntary responses from readers of the USA Today
sports website. Do you think the 2000 respondents actually formed random sample? Explain.
Answer. No – the readership of the website is limited to those who have access to the site
and choose to visit it; moreover, the survey was not based on a random selection of even those
users of the website, but on those who chose to respond to the poll.
14. A recent survey of 104 randomly selected gas stations in California found that the mean
price for unleaded gasoline in the sample was $1.52 per gallon with a sample standard deviation
of $.10.
(a) Conduct an hypothesis test at a 1% level of significance to determine whether the mean
price for unleaded gasoline in California is more than $1.50 per gallon.
(i) State the null and alternative hypotheses.
Answer. The null hypothesis is H0 :
µ > $1.50.
µ = $1.50, and the alternative hypothesis is H1 :
(ii) State or draw the critical region, report the conclusion of your test.
Answer. The critical region is z ≥ 2.33. Now we compute
z=
1.52 − 1.50
√0.1
104
= 2.039
Because z does not fall in the critical region, we conclude that there is not sufficient evidence
to reject the null hypothesis at a level of significance of .01.
(b) Report the Pvalue of the test in (a).
Answer. P (z > 2.04) = .5 − .4793 = .0207
(c) Explain what the Pvalue in (b) means.
Answer. There is about a 98% probability that the true mean price for unleaded gasoline in
California is more that $1.50 per gallon.
(d) If you were to repeat the test in (a) for levels of significance α = .02, α = .03 and α = .05,
what would the conclusion of the test be?
8
Answer. We would reject H0 for levels of significance α = .03 and α = .05 because the Pvalue
is less than these α’s, we would not reject H0 for α = .02.
15. (a) If a population has a standard deviation of 80, what sample size would be necessary
in order for a 95% confidence interval to estimate the population mean within 10?
2
z σ 2
1.96 · 80
c
Answer. We use the formula n =
and we find that n =
= 245.86. The
E
10
sample size must be the next larger whole number which is n = 246.
(b) What size of sample is needed by the Gallup organization to estimate a population proportion within ±.02 with 95% confidence. In your calculation assume that there is no preliminary
estimate for p.
1 zc 2 1
Answer. We use the formula n =
=
4 E
4
1.96
.02
2
which gives us a sample size of 2401.
(c) Suppose you are to construct a 99% confidence interval for the mean using a sample of size
n = 12 from a normal population with unknown standard deviation. What value of tc would
s
you use in the formula x̄ ± tc √ ?
n
Answer. We use a t-distribution with n − 1 = 11 degrees of freedom, and look on the table
in the back cover to find t.99 = 3.106
(d) Find the critical region if a two-tailed test on a mean is conducted using a large sample at
a level of significance of .01.
Answer. With the help of the normal table in the front cover, we find that the critical region
is z ≤ −2.58 or z ≥ 2.58.
(e) Find the critical region for a left-tailed test on a proportion at a level of significance α = .05?
Answer. With the help of the normal table in the front cover, we find that the critical region
is z ≤ −1.645.
(f) Explain what type I and type II errors are in hypothesis tests.
Answer. See text Section 9.1.
(g) What is the probability with which we are willing to risk a type I error called?
Answer. The level of significance, which is denoted by α.
16. (a) A developer wishes to test whether the mean depth of water below the surface in a
large development tract was less than 500 feet. The sample data was as follows: n = 32 test
9
holes, the sample mean was 486 feet, and the standard deviation was s = 53 feet. Complete
the test by computing the Pvalue, and report the conclusion for a 1% level of significance.
Answer: Null Hypothesis: µ = 500
Alternative Hypothesis: µ < 500
Using the sample data, we compute
z=
486 − 500
√53
32
= −1.49
Thus the Pvalue is P (z < −1.49) = .0681 We would not reject the null hypothesis at a level
of significance of .01, because the P-value is larger than 0.01.
(b) What would the conclusion of the test be for a level of significance of α = .05.
Answer. Do not reject H0 since the Pvalue is bigger than .05.
(c) What type of error was possibly made in (b)?
Answer. A type II error.
17. A vendor was concerned that a soft drink machine was not dispensing 6 ounces per cup,
on average. A sample size of 40 gave a mean amount per cup of 5.95 ounces and a standard
deviation of .15 ounce.
(a) Find the Pvalue for an hypothesis test to determine if the mean is different from 6 ounces.
Answer. This is a two-tailed test with
Null Hypothesis: µ = 6
Alternative Hypothesis: µ 6= 6.
Using the data, we compute
z=
5.95 − 6
.15
√
40
= −2.11.
The Pvalue is: P (z < −2.11) + P (z > 2.11) = 2 · P (z < −2.11) = 2(.0174) = .0348
(b) For which of the following levels of significance would the null hypothesis be rejected?
(i) α = .10
(ii) α = .05
(iii ) α = .04
(iv) α = .01
Answer. Reject the null hypothesis in (i), (ii) and (iii) since the Pvalue is smaller than α; do
not reject the null hypothesis in (iv).
(c) For each case in part (b), what type of error has possibly been committed?
10
Answer. Possible Type I error may occur in (i), (ii) and (iii) while a Type II error may occur
in (iv).
(d) Find a 96% confidence interval for the mean amount of soda dispensed per cup.
Answer. For c = .96, zc = 2.05 (approximately), look at z value corresponding to an area of
.98 on table. Thus the confidence interval, using the large sample method (n is at least 30)
yields endpoints:
.15
5.95 ± 2.05 · √
40
and, so the confidence interval is: (5.901, 5.999).
(e) Is your interval in (d) consistent with the test conclusion in (b)(iii)? Explain.
Answer. Yes. In (b)(iii) we have a confidence level of (at least) c = 1 − α = .96 that the
mean is different from 6, while in (d) we had a 96% confidence interval that did not contain
6, and so we were (at least) 96% certain that the mean is different from 6. Notice also the
confidence interval comes very close to containing 6, this is reflected in the Pvalue being very
close to 0.04 in the hypothesis test.
(f) Based on your answer to (b)(iv) would you expect a 99% confidence interval to contain 6?
Answer. Yes, because we were not 99% confident that the mean was different from 6, we
would expect the corresponding confidence interval to contain 6.
(g) Suppose that the population standard deviation is σ = .15, what sample size would be
needed so that the maximum error in a 96% confidence interval is E = .01?
z σ 2
c
Answer. The formula to use is: n =
. So we compute
E
2
2.05 · .15
n=
= 945.5625,
.01
thus we should use a sample size of n = 946.
18. (a) Suppose that a February Gallup poll of 1200 randomly selected voters found that 53
percent support George W. Bush’s energy policy. Conduct an hypothesis test at a level of
significance of α = .01 to test whether the true voter population support for George W. Bush’s
energy policy in February was less than 56 percent.
Answer. The null hypothesis is H0 : p = .56, and the alternative hypothesis is H1 : p < .56.
The critical region is z ≤ −2.33. Since n = 1200 and p = .56 and q = .44, we clearly have
np > 5 and nq > 5. Thus we compute
.53 − .56
≈ −2.09
z=q
(.56)(.44)
1200
11
Because −2.09 does not fall in the critical region, we conclude there is not sufficient evidence
to reject the null hypothesis at a level of significance of 1%.
(b) Report the Pvalue of the test in (a) and give a practical interpretation of it.
Answer. The Pvalue is P (z < −2.09) = .0183. Thus we are roughly 98% certain that less
than 56% of all voters at the time of the poll supported President G.W. Bush’s energy policy.
19. (From p. 536 #8) A reading test is given to both a control group and an experimental
group (which received special tutoring). The average score for the 30 subjects in the control
group was 349.2 with a standard deviation of 56.6. The average score for the 30 subjects in the
experimental group was 368.4 with a standard deviation of 39.5. Use a 4% level of significance
to test the claim that the experimental group performed better than the control group.
Answer. We wish to conduct the test H0 : µ1 = µ2 versus H1 : µ1 > µ2 where µ1 is
the population mean test score for all people whosreceived the special tutoring. We use the
(x̄1 − x̄2 ) − (µ1 − µ2 )
σ12 σ22
where σx̄−x̄2 =
+
Thus
σx̄1 −x̄2
n1 n2
r
39.52 56.62
σx̄−x̄2 =
+
= 12.6013
30
30
(368.4 − 349.2) − 0
Therefore, z =
= 1.52. The Pvalue is P (z > 1.52) = 1 − .9357 = 0.0643
12.6013
Because the Pvalue is larger than α = .04, we do not reject the null hypothesis at a 4% level
of significance. There is not sufficient evidence to show that the tutoring increases the mean
score.
formula z =
20. (From p. 505 #14) Nationally about 28% of the population believes that NAFTA benefits
America. A random sample of 48 interstate truck drivers showed that 19 believe NAFTA
benefits America. Conduct an hypothesis test to determine whether the population proportion
of interstate truckers who believe NAFTA benefits America is higher than 28%. Test at a level
of significance of α = .05.
(a) State the null and alternative hypotheses. Is this a right-tailed, left-tailed or two-tailed
test?
Answer. The null hypothesis is H0 : p = .28, the alternative hypothesis is H1 : p > .28. This
is a right-tailed test.
(b) Find the Pvalue for the test.
Answer. For this data, we have n = 48, p = .28 and q = .72. Clearly np > 5 and nq > 5 and
19
so we compute p̂ =
= .39583; then
48
.39583 − .28
≈ 1.79
z= q
(0.28)(0.72)
48
12
Therefore, the Pvalue is P (z > 1.79) = 1 − .9633 = .0367. Because the Pvalue is less than
α = .05, we reject H0 . The results are statistically significant; they indicate that the proportion
of interstate truck drivers that believe NAFTA benefits America is higher than 0.28.
(c) Would you reject the null hypothesis at a level of significance of α = .04?
Answer. Yes, because the Pvalue is less than 0.04.
(d) Would you reject the null hypothesis at a level of significance of α = .01?
Answer. No, because the Pvalue is more than 0.01.
(e) Do you think the proportion of interstate truckers who believe NAFTA benefits America
is higher than 28%? Explain your answer.
Answer. Yes, and I’m quite sure of this, in fact, I’m roughly 96% sure of it.
21. (From p. 539 #18) A random sample of n1 = 288 voters registered in the state of California
showed that 141 voted in the last general election. A random sample of n2 = 216 registered
voters in Colorado showed that 125 voted in the last general election. Do these data indicate
that the population proportion of voter turnout in Colorado is higher than that in California?
Use a 5% level of significance.
Answer. Let p1 and p2 represent the population proportions of voter turnout in California and
Colorado respectively. We will test H0 : p1 = p2 versus H1 : p1 < p2 . Since we are assuming
p1 = p2 , we use the pooled estimate for proportions (see p. 530), that is
p̂ =
141 + 125
= .5278.
288 + 216
Then we use this for an estimate of p1 and p2 in the formula for σp̂1 −p̂2 , so we obtain
r
(0.5278)(0.4722) (0.5278)(0.4722)
σp̂1 −p̂2 =
+
= 0.04494.
288
216
From this, we compute
z=
(p̂1 − p̂2 ) − (p1 − p2 )
(.4896 − .5787) − 0
=
= −1.98.
σp̂1 −p̂2
0.04494
The Pvalue is P (z < −1.98) = 0.0239. Because the Pvalue is less than .05, we reject H0 at the
5% level of significance.
13