Download Dear niramol, Great questions! Here`s how I`d answer them: These

Dear niramol,
Great questions! Here's how I'd answer them:
These are my questions:
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.
The test statistic is z=(9-10)/(2.8/sqrt(50) = -2.525
From the normal table, we find: P(z<-2.525) = 0.00578.
That's the pvalue.
Hence there is very strong evidence that people in the
program lose fewer 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
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.
Here we use a z-test because we're given the standard
deviation of the population.
Hypotheses:
H0: mu<=16
Ha: mu>16
The test statistic is:
z=(16.05-16)/(0.03/sqrt(50))= 11.785
And from the z-table, we find the p-value to be less than
0.0001. So we reject the null hypothesis. The mean is
greater than 16.
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.
In this case:
H0: mu>= 6
Ha: mu<6
The sample statistics is:
average: 5.6375
standard deviation: 0.635
The test statistic is
t=(5.6375-6)/(0.635/sqrt(8)) = -1.615
We get the p-value from the t-table or from using excel.
The excel formula is: =tdist(1.615, 7, 1) = 0.075
So we do not reject the null hypothesis at the 0.05 level.
There is not enough evidence to suggest that the mean
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.
27.
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?
We are given the population standard deviations, so we
need to use a z-test.
Here is the correct answer:
Statistic
p;
Men &nbs
Women
Sample mean
& nbsp;
24.51
;
22.69
Population standard deviation &nb
sp;
4.48
;
3.86
Sample size
35
;
&nbs p;
40
Hypotheses:
H0: mu1 = mu2
Ha: mu1 not equal to mu2
The test statistic is:
z=(24.51 - 22.69)/sqrt(4.48^2/35+3.86^2/40) = 1.871
So we find the p-value to be 0.0613
So we do not reject the null hypothesis. There is not
enough evidence to suggest that the means are equal at
the .01 level of significance.
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:
27.
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?
The sample statistics are:
Little River Mean: 27.461
Little River Standard deviation: 4.44
Little Rivver Variance: 19.717
Murrells Inlet Mean: 25.689
Murrells Inlet Standard deviation: 2.685
Murrells Inlet Variance: 7.207
So our hypotheses are:
H0: mu1 = mu2
Ha: mu1 not equal to mu2
The critical value is calculated like this:
Since we're told not to assume equal variance, we have
to calculate the degrees of freedom. The formula is:
df = (19.727/10 + 7.207/12)^2/((19.717/10)^2/9 +
(7.207/12)^2/11) = 14.248
So we'll use 15 degrees of freedom.
So we go to the t-table and get the critical value. It's
2.131.
The test statistic is:
t=(27.461-25.689)/sqrt(19.717/10+7.207/12) = 1.105
So we do not reject the null hypothesis.
There is not enough evidence to suggest that the means
are not the same.
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?
Family
Progressive Car Insurance
GEICO 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
We'll work a paired t-test for this one. So we need the
column of differences:
It is:
480
436
-925
463
-531
-1,092
120
-785
-938
109
196
-653
-188
-59
-328
And we need the average and standard deviation:
The average difference is -246
The standard deviation is 546.959
So we just work a one sample t-test now:
The hypotheses:
H0: mu1-mu2 = 0
Ha: mu1-mu2 not equal to 0
The critical value is 1.761 and -1.761. That's with 14
degrees of freedom.
The test statistic is:
t=-246/(546.959/sqrt(15)) = -1.742
So we do not reject the null hypothesis. There's not
enough evidence to say that the means are unequal.
Chapter 12 ex 23, 28
23. 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?
Hypotheses:
H0: sigma_oceanfront <= sigma_threeblocks
Ha: sigma_oceanfront > sigma_threeblocks
The critical value is from the F-table. The degrees of
freedom are 20 and 17. The value is 3.16
The test statistic is:
F = 45600^2 / 21330^2 = 4.570
So we reject the null hypothesis. The variance in price of
oceanfront homes is greater than that of homes 3 blocks
away.
28. The following is a partial ANOVA table.
23.
Source
Sum of Squares
df
Mean Square
F
Treatment
2
Error
20
Total
500
11
Complete the table and answer the following questions.
Use the .05 significance level.
I'll give you the ANOVA table one column at a time:
Sum of Squares
320
180
500
Degrees of Freedom
2
9
11
Mean Square
160
20
F-Value
8
--I Found the questions in another post. Here they are:
a.
How many treatments are there?
There are 3 treatments. We know this because the
degrees of freedom for treatments is 2.
b.
What is the total sample size?
The total sample size is 12. That's because the degrees
of freedom for total is equal to 11.
c.
What is the critical value of F?
We get this from the F-table. The degrees of freedom
are 2 and 9. The critical value is 4.256
d.
Write out the null and alternate hypotheses.
H0: mu1=mu2=mu3
Ha: not H0. That is, at least one of the means is
different.
e.
What is your conclusion regarding the null
hypothesis?
We reject the null hypothesis. The means are not all
equal. At least one is different.
Chapter 17 ex 19, 20
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: The proportions are the same
Ha: The proportions are not the same.
The degrees of freedom is equal to 2, so the critical
value, from the chi-square table, is 5.991
Chi square = (53-50)^2 /50 + (64-50)^2 /50 + (3350)^2 /50 = 9.88
So we reject the null hypothesis. The proportions are not
the same.
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?
17.
Entrance
Frequency
Main Street
140
Broad Street
120
Cherry Street
90
Walnut Street
50
Total
400
Hypotheses:
H0: p1 = p2 = p3 = p4 = 1/4
Ha: not H0.
The degrees of freedom is 4-1 = 3. So the critical value
is gotten from the chi-square table. It is 11.345.
The test statistic is:
X2 = (140-100)^2/100 + (120-100)^2/100 + (90100)^2/100 + (50-100)^2/100 = 46
So we reject the null hypothesis. The entrances are not
equally utilized!.
---Let me know if you have questions. I'm happy to clarify
any details!