Ww2.justanswer.com
CHAPTER 10
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.
H0: mean = 10 pounds
Ha: mean < 10 pounds
critical value is -1.6449 from a table
z = (x-mu)/(sigma/sqrt(N))
z = (9-10)/(2.8/sqrt(50))
z = -2.52538
The p value is about 0.006 (from a table)
The z value is lower than the critical value, so we reject the null, and conclude that they lose less than 10 pounds.
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 qualitycontrol
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
H0: mean = 16
Ha: mean > 16
Get the test statistic:
z = (x-mu)/(sigma/sqrt(N))
z = (16.05-16)/(0.03/sqrt(50))
z = 11.785113
Using a p value calculator:
The p value is so slightly above 0 that the calculator shows 0.
Since the p value is so small, there is enough evidence to suggest that the cans are over filled.
. 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 =
|5.6375 |
stdev =
|0.63457 |
H0: mean rate = 6
Ha: mean rate < 6
Get the t statistic:
t = (x-mu)/(sigma/sqrt(N))
t = (5.6375-6)/(0.63457/sqrt(8))
t = -1.615747
The p value, one tail, df = N-1 = 7, is:
0.075064
This p value is greater than the 0.01 significance, so we do not reject the null hypothesis. There is not enough evidence to conclude that the rate is less than 6.
CHAPTER 11
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 Men Women
Sample mean 24.51 22.69
Population standard deviation 4.48 3.86
Sample size 35 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?
H0: men mean – women mean = 0
Ha: men mean – women mean not equal to 0
Get the test statistic:
(24.51-22.69)/sqrt(4.48^2/35 + 3.86^2/40)
= 1.871
The p value (from a table) is:
About 0.06
This is well above the 0.01 significance, so we do not reject the null hypothesis. There is no statistic difference.
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 Waiting Time
Little River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49
Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 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?
See Excel file
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 GEICO
Family Car Insurance Mutual Insurance
Becker $2,090 $1,610
Berry 1,683 1,247
Cobb 1,402 2,327
Debuck 1,830 1,367
DuBrul 930 1,461
Eckroate 697 1,789
German 1,741 1,621
Glasson 1,129 1,914
King 1,018 1,956
Kucic 1,881 1,772
Meredith 1,571 1,375
Obeid 874 1,527
Price 1,579 1,767
Phillips 1,577 1,636
Tresize 860 1,188
See Excel File!
CHAPTER 12
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?
H0: variances are equal
Ha: variances are not equal
dfN = 21-1 = 20
dfD = 18-1 = 17
Using a table, the critical F value, with 0.01 significance is:
3.1615
Get the test statistic:
sigma1^2 / sigma2^2
= 45600^2/21330^2
= 4.57033
The test statistic is greater than the critical value, so we reject the null hypothesis and conclude that the variances are different.
28. The following is a partial ANOVA table.
Sum of Mean
Source Squares df Square F
____________________________________________________
Treatment 2
Error 20
Total 500 11
|Source |df |Sum of squares |Mean square |F |
|Treatments | 2 | 320 |160 | 8 |
|Error | 9 | 180 |20 | |
|Total | 11 | 500 | | |
Complete the table and answer the following questions. Use the .05 significance level.
a. How many treatments are there?
2+1 = 3
b. What is the total sample size?
11+1 = 12
c. What is the critical value of F?
4.2565 (from a table)
d. Write out the null and alternate hypotheses.
H0: means are equal
Ha: means are not equal
e. What is your conclusion regarding the null hypothesis?
Since our F statistic (8) is greater than the critical value, we reject the null. The means are different.
CHAPTER 17
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?
H0: no difference between channels
Ha: difference between channels
Df = n-1 = 3-1 = 2
The critical value, from a table, is 5.991
Chi square = (53-50)^2 /50 + (64-50)^2 /50 + (33-50)^2 /50 = 9.88
The statistic is much higher than the critical value, so we reject the null hypothesis. There is a difference between the channels.
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 Frequency
Main Street 140
Broad Street 120
Cherry Street 90
Walnut Street 50
Total 400
H0: no difference between the entrances
Ha: difference between the entrances
Using a table (), the critical value is (df = N-1 = 3, alpha = 0.01):
11.345
So we reject the null for values greater than that.
The expected value for each entrance is 100, so get chi squared:
(140-100)^2/100 + (120-100)^2/100 + (90-100)^2/100 + (50-100)^2/100
= 46
The chi squared value is much higher than the critical value, so we reject the null hypothesis and conclude that there is a difference between the entrances.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.