Download 31 - JustAnswer

31. A new weight-watching company, Weight Reducers International, advertises that those
who join will lose, on the average, 10 pounds the first two weeks with a standard deviation
of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program
revealed the mean loss to be 9 pounds. At the .05 level of significance, can we
conclude that those joining Weight Reducers on average will lose less than 10 pounds?
Determine the p-value.
Hypotheses:
H0: µ ≥ 10
Ha: µ < 10 (Claim)
Critical value:
For a left-sided, one-tailed test with
= 0.05, z(crit) = -1.645
Test value:
z(test) = (9 – 10) / (2.8 / √50) = -2.5254
p-value = 0.0058
Decision:
z(test) < z(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.05 level of significance to support the
claim that those joining Weight Reducers will lose less than 10 pounds
on average.
32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being
overfilled. Assume the standard deviation of the process is .03 ounces. The quality control
department took a random sample of 50 cans and found that the arithmetic mean
weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that
the mean weight is greater than 16 ounces? Determine the p-value.
Hypotheses:
H0: µ ≤ 16
Ha: µ > 16 (Claim)
Critical value:
For a right-sided, one-tailed test with
= 0.05, z(crit) = 1.645
Test value:
z(test) = (16.05 – 16) / (0.03 / √50) = 11.7851
p-value = 0.00 (when rounded off)
Decision:
z(test) > z(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.05 level of significance to support the
claim that the mean weight is greater than 16 ounces.
38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now
less than 6 percent. A sample of eight small banks in the Midwest revealed the following
30-year rates (in percent):
4.8
5.3
6.5
4.8
6.1
5.8 6.2
5.6
At the .01 significance level, can we conclude that the 30-year mortgage rate for small
banks is less than 6 percent? Estimate the p-value.
Mean: x-bar = (4.8 + 5.3 + 6.5 + 4.8 + 6.1 + 5.8 + 6.2 + 5.6) / 8 = 5.6375
Standard Deviation: s = 0.6346
Hypotheses:
H0: µ ≥ 6
Ha: µ < 6 (Claim)
Critical value:
For a left-sided, one-tailed test with
t(crit) = -2.998
= 0.01, and 7 degrees of freedom,
Test value:
z(test) = (5.6375 – 6) / (0.6346 / √8) = -1.6157
p-value = 0.0751
Decision:
z(test) > z(crit)  Do not reject the null hypothesis
Summary:
There is not sufficient evidence at the 0.01 level of significance to support the
claim that the mean 30-year mortgage rate for small banks is less than 6%.
27. A recent study focused on the number of times men and women who live alone buy
take-out dinner in a month. The information is summarized below.
Statistic
Sample mean
Population standard deviation
Sample size
Men
24.51
3.86
35
4.48
Women
22.69
40
At the .01 significance level, is there a difference in the mean number of times men and
women order take-out dinners in a month? What is the p-value?
Hypotheses:
H0: µ1 = µ2
Ha: µ1 ≠ µ2 (Claim)
Critical value:
For a two-tailed test with
= 0.01, z(crit) = ±2.576
Test value:
ztest 
x1  x 2 
2
1
2
s
s
 2
n1 n 2

24.51  22.69
4.48 2 3.86 2

35
40
 1.8713
p-value = 0.0613
 Decision:
z(test) < z(crit)  Do not reject the null hypothesis
Summary:
There is not sufficient evidence at the 0.01 level of significance to support the
claim that there is a difference in the mean number of times men and women
order take-out dinners in a month.
46. Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies
for visitors to the Myrtle Beach area. There are two facilities, one in the Little
River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to
compare the mean waiting time for patients at the two locations. Samples of the waiting
times, reported in minutes, follow:
Location
Little River
Murrells Inlet
Waiting Time
28.77 29.53 22.08 29.47
31.73
22.93
23.92
26.92
27.20 26.44
25.62
18.60
30.61
32.94
29.44
25.18
23.09
29.82
23.10
26.49
26.69
22.31
Assume the population standard deviations are not the same. At the .05 significance level,
is there a difference in the mean waiting time?
Little River :
x1  27.5689
s1  4.6959
n1  9
Murrels Inlet :
x2  25.9270
s2  2.6788
n 2  12
Hypotheses:

H0: µ1 = µ2
Ha: µ1 ≠ µ2 (Claim)
Critical value:
For a two-tailed test with
= 0.05 and 8 degrees of freedom, t(crit) = ± 2.306
Test value:
t test 
x1  x 2 
2
1
2
s
s
 2
n1 n 2

27.5689  25.9270
4.6959 2 2.6788 2

9
12
 0.9404
p-value = 0.3745

Decision:
t(test) < t(crit)  Do not reject the null hypothesis
Summary:
There is not sufficient evidence at the 0.05 level of significance to support the
claim that there is a difference in the mean waiting time.
52. The president of the American Insurance Institute wants to compare the yearly costs of
auto insurance offered by two leading companies. He selects a sample of 15 families,
some with only a single insured driver, others with several teenage drivers, and pays each
family a stipend to contact the two companies and ask for a price quote. To make the
data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is
reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?
Progressive
Car Insurance
$2,090
1,683
1,402
1,830
930
697
1,741
1,129
1,018
1,881
1,571
874
1,579
1,577
860
Family
Becker
Berry
Cobb
Debuck
DuBrul
Eckroate
German
Glasson
King
Kucic
Meredith
Obeid
Price
Phillips
Tresize
Progressive :
x1  1391.00
s1  437.7562
n1  15
GEICO
Mutual Insurance
$1,610
1,247
2,327
1,367
1,461
1,789
1,621
1,914
1,956
1,772
1,375
1,527
1,767
1,636
1,188
Geico :
x2  1637
s2  299.0611
n 2  15
Hypotheses:

H0: µ1 = µ2
Ha: µ1 ≠ µ2 (Claim)
Critical value:
Rather than testing the variances to determine if they are equal or unequal, which
is controversial according to some statisticians, simply calculate both:
Assuming that the variances are unequal:
For a two-tailed test with
= 0.10 and 14 degrees of freedom, t(crit) = ±1.761
Assuming that the variances are equal:
For a two-tailed test with
= 0.10 and 28 degrees of freedom, t(crit) = ±1.701
Test value:
Variances unequal:
t test 
x1  x 2 
s12 s2 2

n1 n 2

1391 1637
437.7562 2 299.06112

15
15
 1.7971
p-value = 0.0939

Variances equal:
t test 
x1  x 2 
n1 1s12  n2 1s22
n1  n 2  2
t(test) 
1 1

n1 n 2
1391 1637
14437.75622  14 299.06112 1
15 15  2
 1.7971
1

15 15
p-value = 0.0831

Decision:
Variances unequal:
t(test) < t(crit)  Reject the null hypothesis
Variances equal:
t(test) < t(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.10 level of significance to support the claim
that there is a difference in the amounts quoted.
23. A real estate agent in the coastal area of Georgia wants to compare the variation in the
selling price of homes on the oceanfront with those one to three blocks from the ocean.
A sample of 21 oceanfront homes sold within the last year revealed the standard deviation
of the selling prices was $45,600. A sample of 18 homes, also sold within the last
year, that were one to three blocks from the ocean revealed that the standard deviation
was $21,330. At the .01 significance level, can we conclude that there is more variation
in the selling prices of the oceanfront homes?
Use the F-test to compare variances.
The sample with the greater variance is designated as “sample 1” and this variance goes
in the numerator of the calculated F-statistic.
In this scenario, the oceanfront homes are designated as “sample 1”.
Hypotheses:
H0: s12 ≤ s22
Ha: s12 > s22 (Claim)
Critical value:
For a right-sided one-tailed test with
and n2 = 18 – 1 = 17
= 0.01, n1 = 21 – 1 = 20,
F(crit) = 3.16
Test value:
F test 
s12 45,600

 4.5703
s2 2 21,3302
2
Decision:

F(test) > F(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.01 level of significance to support the claim
that there is more variation in the selling prices of the oceanfront homes.
28. The following is a partial ANOVA table.
Source
Treatment
Error
Total
Sum of
Squares
df
2
Mean
Square
F
20
500
11
Complete the table and answer the following questions. Use the .05 significance level.
The completed chart:
Source
Treatment
Error
Total
Sum of
Squares
320
180
500
df
2
9
11
Mean
Square
160
20
F
8
a. How many treatments are there?
There are 3 treatments
b. What is the total sample size?
The total sample size is 12
c. What is the critical value of F?
The critical value of F is 4.26
d. Write out the null and alternate hypotheses.
H0: µ1 = µ2 = µ3
Ha: At least one of the means is different from the others. (Claim)
e. What is your conclusion regarding the null hypothesis?
Since F(test) is greater than the critical value, reject the null hypothesis.
19. In a particular market there are three commercial television stations, each with its own
evening news program from 6:00 to 6:30 P.M. According to a report in this morning’s local
newspaper, a random sample of 150 viewers last night revealed 53 watched the news
on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13).
At the .05 significance level, is there a difference in the proportion of viewers watching
the three channels?
If the proportions were all equal between the three tv channels, then they would each be
equal to 1/3.
Hypotheses:
H0: p1 = p2 = p3 = 1/3
Ha: At least one proportion is different from 1/3 (Claim)
Critical value:
Using a Chi-Squared Test, the critical
of freedom, at a = 0.05 is:
2
2
value for k – 1 = 3 – 1 = 2 degrees
(crit) = 5.991
Test value:
The expected number of viewer of each channel, if the proportions are equal is:
np = (150)(1/3) = 50
The
2
test statistic is:
2
(test) = (53 – 50)2 / 50 + (64 – 50)2 / 50 + (33 – 50)2 / 50 = 9.88
Decision:
2
(test) >
2
(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.05 level of significance to support the claim
that there is a difference in the proportion of viewers watching each channel.
20. There are four entrances to the Government Center Building in downtown Philadelphia.
The building maintenance supervisor would like to know if the entrances are equally utilized.
To investigate, 400 people were observed entering the building. The number using
each entrance is reported below. At the .01 significance level, is there a difference in the
use of the four entrances?
Entrance
Main Street
Broad Street
Cherry Street
Walnut Street
Total
Frequency
140
120
90
50
400
As in the previous problem, if the use of all four entrances was the same, then the
proportion using each door would be the same. The expected value for each door should
then be 400 /4 = 100
Hypotheses:
H0: p1 = p2 = p3 = p4
Ha: At least one proportion is different from 1/4 (Claim)
Critical value:
Using a Chi-Squared Test, the critical
freedom, at
2
2
value for k – 1 = 4 – 1 = 3 degrees of
= 0.01 is:
(crit) = 11.345
Test value:
The
2
test statistic is:
2
(test) = (140 – 100)2 / 100 + (120 – 100)2 / 100 + (90 – 100)2 / 100
+ (50 – 100)2 / 100 = 46
2
(test) >
Decision:
2
(crit)  Reject the null hypothesis
Summary:
There is sufficient evidence at the 0.05 level of significance to support the claim
that there is a difference in the use of the four entrances.