A sample of 9 days over the past six months showed that a clinic treated the following numbers of patients: 24, 26, 21, 17, 16, 23, 27, 18, and 25. If the number of patients seen per day is normally...

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A sample of 9 days over the past six months showed that a clinic treated the following numbers of
patients: 24, 26, 21, 17, 16, 23, 27, 18, and 25. If the number of patients seen per day is normally
distributed, would an analysis of these sample data provide evidence that the variance in the number
of patients seen per day is less than 10? Use a = .025 level of significance. What is your conclusion
using p-value and critical value approaches. Is the conclusion different in both the cases?
2. At Western University the historical mean of scholarship examination score for freshman applications
is 900. Population standard deviation is assumed to be known as 180. Each year, the assistant dean
uses a sample of applications to determine whether the mean examination score for the new freshman
applications has changed.
a) State the hypotheses.
b) What is the 95% confidence interval estimate of the population mean examination score if a
sample of 100 applications provided a sample mean 935?
c) Use the confidence interval approach to conduct a hypothesis test. What is your conclusion?
d) Assuming a = .05, conduct p-value based and critical-value based hypothesis tests. How do the
results compare in all the three cases?
3. The grade point averages of 61 students who completed a college course in financial accounting have
a standard deviation of .790. The grade point averages of 17 students who dropped out of the same
course have a standard deviation of .940. Do the data indicate a difference between the variances of
grade point averages for students who completed a financial accounting course and students who
dropped out? Use a = .05 level of significance. Use both p-value and critical value approaches.
Compare the test results.
4. The following data were collected on the number of emergency ambulance calls for an urban county
and a rural county in Florida. Is County type independent of the day of the week in receiving the
emergency ambulance calls? Use a = 0.005. What is your conclusion?
Day of the Week
Sun Mon Tue Wed Thu Fri Sat
County
Urban 62 47 48 51 61 74 41
Rural 6 10 18 17 11 13 12
5. Medical tests were conducted to learn about drug-resistant tuberculosis. Of 284 cases tested in New
Jersey, 18 were found to be drug- resistant. Of 536 cases tested in Texas, 10 were found to be drugresistant.
Do these data indicate that New Jersey has a statistically significant higher outbreak of drugresistant
tuberculosis cases? Use a .03 level of significance. What is the p- value, and what is your
conclusion? Is the conclusion any different under critical-value approach?
6. Consider the following data for two independent random samples taken from two normal populations.
Sample 1 14 26 20 16 14 18
Sample 2 18 16 8 12 16 14
a) Compute the two sample means and the two sample standard deviations.
b) What is the point estimate of the difference between the two population means?
c) Assuming a = .10, conduct p-value based and critical-value based hypothesis tests for the equality
of means of the two populations.
d) What is the 90% confidence interval estimate of the difference between the two population means?
How do the results compare in all the three approaches to hypothesis testing?

Homework #8


Chapter 10

1. The Professional Golf Association (PGA) measured the putting accuracy of professional
golfers playing on the PGA Tour and the best amateur golfers playing in the World Amateur
Championship. A sample of 1800 6-foot putts by amateur golfers found 1044 made putts. A
sample of 1700 6-foot putts by professional golfers found 1088 made puts.
a. Estimate the proportion of made 6-foot putts by professional golfers. Estimate the
proportion of made 6-foot putts by amateur golfers. Which group had a better putting
accuracy?
b. What is the point estimate of the difference between the proportions of the two
populations? What does this estimate tell you about the percentage of putts made by the
two groups of golfers?
c. What is the 97% confidence interval for the difference between the two population
proportions? Interpret his confidence interval in terms of the percentage of putts made by
the two groups of golfers.
2. A May 1993 New York Times/CBS News poll found that of 500 adults who were planning a
vacation in the next six months, 85 were expecting to travel by airplane. A similar survey
question in a 2003 New York Times/CBS News poll sampled 700 adults who were planning
a vacation during the next six months and found that 189 were expecting to travel by
airplane.
a. State the hypotheses that can be used to determine whether a significant change occurred
in the population proportion planning to travel by airplane over the 10-year period.
b. What is the sample proportion expecting to travel by airplane in 2003? In 1993?
c. Use a = .06 and test for a significant difference. What is your conclusion?
d. Discuss reasons that might provide an explanation for this conclusion.
2

Chapter 11

3. Find the following chi-square distribution values.
a. ?
2
0.005 with Sample Size = 101
b. ?
2
0.10 with Sample Size = 30
c. ?
2
0.05 with Sample Size = 16
d. ?
2
0.995 with Sample Size = 19
e. ?
2
0.99 with Sample Size = 96
4. A sample of 27 items provides a sample standard deviation of 10. Test the following
hypotheses using a = .10. What is your conclusion? Use both the p-value approach and the
critical value approach.
H0: s
2
= 75
Ha: s
2 > 75
5. The daily car rental rates for a sample of eight cities follow.
a. What is the 98% confidence interval estimate of the variance of car rental rates for the
population?
b. What is the 99% confidence interval estimate of the standard deviation for the
population?
3
6. At the end of 2008, the variance in the semiannual yields of overseas government bond was
s
2 = 0.25. A group of bond investors met at that time to discuss future trends in overseas
bond yields. Some expected the variability in overseas bond yields to increase and others
took the opposite view. The following table shows the semiannual yields for 12 overseas
countries as of March 6, 2009.
a. Compute the mean, variance, and standard deviation of the overseas bond yields as of
March 6, 2009.
b. Develop hypotheses to test whether the sample data indicate that the variance in bond
yields has increased from that at the end of 2008.
c. Use a = .025 to conduct the hypothesis test formulated in part (b). What is your
conclusion?
7. Find the critical values based on F distribution when variance of sample 1 is greater than
variance of sample 2. Alpha and Sample Sizes are given below.
a. a = .05 with n1 = 6 and n2 = 11
b. a = .025 with n1 = 21 and n2 = 26
c. a = .01 with n1 = 61 and n2 = 61
d. a = .10 with n1 = 9 and n2 = 25
4
8. The variance in a production process is an important measure of the quality of the process. A
large variance often signals an opportunity for improvement in the process by finding ways
to reduce the process variance. Conduct a statistical test to determine whether there is a
significant difference between the variances in the bag weights for two machines. Use a .02
level of significance. What is your conclusion? Which machine of the two provides the
greater opportunity for quality improvements?
9. Fidelity Magellan is a large cap growth mutual fund and Fidelity Small Cap Stock is a small
cap growth mutual fund. For Fidelity Magellan, the sample standard deviation is 18.89; for
Fidelity Small Cap Stock, the sample standard deviation is 13.03. The standard deviation for
both funds was computed based on a sample of sizes 26 and 25 respectively. Financial
analysts often use the standard deviation as a measure of risk. Conduct a hypothesis test to
determine whether the large cap growth fund is riskier than the smaller cap growth fund. Use
a = .10 as the level of significance.
3.46

Homework #9

1. Data from the U.S. Shopper Database provided the following percentages for women shopping
at each of the various outlets. The other category included outlets such as Target, Kmart, and
Sears as well as numerous smaller specialty outlets. No individual outlet in this group accounted
for more than 5% of the women shoppers. A recent survey using a sample of 200 women
shoppers in Tampa, Florida, found 60 Wal-Mart, 29 traditional department store, 11 JC Penney,
14 Kohl’s, 30 mail order, and 56 other outlet shoppers. Does this sample suggest that women
shoppers in Tampa differ from the preferences expressed in the U.S. Shopper Data-base? What is
your conclusion based on both the p-value and critical-value approaches? Use a = .01.

Outlet Percentage

Other 35
Wal-Mart 25
Department Stores 10
Mail Order 15
Kohl's 10
J.C. Penney 5
2. The Wall Street Journal’s Shareholder Scoreboard tracks the performance of 1000 largest U.S.
companies. The performance of each company is rated based on the annual total return, including
stock price changes and the re-investment of dividends. Ratings are assigned by dividing all
1000 largest U.S. companies into four groups of equal size Group A (top rating), B (second best
rating), C (third best rating), and D (bottom most rating). Shown here are the one- year ratings
for a sample of 50 largest U.S. companies. Does the sample data provide evidence that the
ratings are equally likely for the largest U.S. companies? Use a = .025.

A B C D

22 9 14 5
3. With double- digit annual percentage increases in the cost of health insurance, more and more
workers are likely to lack health insurance coverage. The following sample data provide a
comparison of workers with and without health insurance coverage for small, medium, and large
companies. For the purposes of this study, small companies are companies that have fewer than
100 employees. Medium companies have 100 to 999 employees, and large companies have 1000
or more employees. Sample data is reported as follows:

Health Insurance


Size of Company Yes No Total

Small 50 25 75
Medium 80 20 100
Large 115 10 125
Total 245 55 300
a. Conduct a test of independence using critical value approach to determine whether
employee health insurance coverage is independent of the size of the company. Use a =
.005.
b. What is the p- value, and what is your conclusion?
c. The USA Today article indicated employees of small companies are more likely to lack
health insurance coverage. Use percentages based on the preceding data to support this
conclusion.
4. FlightStats, Inc., collects data on the number of flights scheduled and the number of flights
flown at major airports throughout the United States. FlightStats data showed 56% of flights
scheduled at Newark, La Guardia, and Kennedy airports were flown during a three-day
snowstorm. All airlines say they always operate within set safety parameters— if conditions are
too poor, they don’t fly. The following data show a sample of 600 scheduled flights during the
snowstorm. Use the chi- square test with a .10 level of significance to determine whether or not
flying/ not flying in a snowstorm is independent of Airliner. What is your conclusion based on
Critical Value test? Is it any different from conclusion based on a p-value approach?

Flight American Continental Delta United

Yes 70 105 95 45
No 80 55 85 65
5. The number of incoming phone calls defined by a Random Variable X at a company
switchboard during 1- minute intervals is believed to have a Poisson distribution. Use a .05 level
of significance and the following data to test the assumption that the incoming phone calls follow
a Poisson distribution.


x


Observed Freq.

0 14
1 33
2 48
3 44
4 30
5 15
6 9
7 6
8 1
Answered Same DayDec 21, 2021

Answer To: A sample of 9 days over the past six months showed that a clinic treated the following numbers of...

David answered on Dec 21 2021
124 Votes
Order ID: TTs161112_58288_1
Assignment Title: Please Solve,
Homework 7
Answer 1) This is a small sample problem testing of mean. Parent population is normal and
population mean is not known. Hence we use chi-square test
For testing H0: σ
2≥10
Against H1: σ
2<10 test statistics is given by
 
2
2 2
;2
i
i
n
x


 





Here is μ population mean which is unknown so we replace it by its unbiased estimate sample
mean which is obtained from sample.
Unbiased estimate of μ =
1
1 n
i
i
x x
n 
 
Hence now the test statistics
under H0 is given by
 
2
2 2
; 12
0
i
i
n
x x
 






Where σ
2
0 is the value we want to test, in our case it is 10.
Now for calculation of test statistics we make following table
Serial
no
xi xi-21.88 (xi-21.88)
2
1 24 2.12 4.4944
2 26 4.12 16.9744
3 21 -0.88 0.7744
4 17 -4.88 23.8144
5 16 -5.88 34.5744
6 23 1.12 1.2544
7 27 5.12 26.2144
8 18 -3.88 15.0544
9 25 3.12 9.7344
Total 197 0.08 132.8896
We have n=9 (sample size)
24 26 21 17 16 23 27 18 25
9
197 / 9 21.88
x
x
        
  
 
 

Hence we have test statistics as
2 2
.025,8
132.8896
13.28896
10
   this is our calculated value.
Now we have a left sided tail test so we check value of corresponding to 8 degree of freedom
from the chi square table for 0.025 level of significance.
It is
2
8;.025 2.18  . This gives us our tabulated value.
Now our calculated value is more than tabulated value, we fail to reject H0. That is these sample
data do not provide evidence that the variance in the number of patients seen per day is less than
10. P value is area corresponding to chi square value of test statistics=13.28896 from table we
the value =0.051
Here p> level of significance So we reject null hypothesis. Result from both approach is same.
Answer 2) it is a large sample problem and testing of mean for known variance(population
variance is given)μ=900 and σ=180
a) Null hypotheses for given problem is
H0: μ=900 against the null hypothesis
H1: μ≠900
b) Level of significance is 0.05
Test statistic is given by
(0,1)
x
z N
n




We are given sample mean=935 and n=100
Under H0 test statistics is given by
935 900
(0,1)
180
100
z N


Z=35/18=1.94
Confidence interval can be found as follows
 
 
 
 
 
 
1.96 .05
1.96 .95
1.96 .95
1.96 1.96 .95
1.96(18) 1.96(18) .95
35.28 35.28 .95
935 35.28 935 35.28 .95
899.72 970.28 .95
P z
P z
x
P
n
P x x
n n
P x x
P x x
P
P


 





 
 
 
 
  
 
 
 
     
 
    
    
    
  

Hence 95% confidence interval estimate of population mean is (899.72, 970.28)
c) We say that (899.72, 970.28) is the 95% confidence limit that is 95 % of times population
mean would fall in this. Value of population mean specified by null hypothesis= 900 falls
in this interval so we accept null at .05 level of significance.
d) At 0 .05 level of significance our test statistics is 1.94 and tabulated value of z is 1.96.
Here calculated value < tabulated value hence we fail to reject null hypothesis on level of
significance basis.
p-value= 2*area to left of tabulated z

 2 1.94
0.0524
P z 


Here p-value > α hence we accept null hypothesis.
So interval estimate and p-value ask to accept null hypothesis while level of significance
asks to reject null hypothesis.
Answer 3) it is a problem for testing equality of variance. So we use F-test for testing
Here we have n1=61, n2=17,
s1
2
= (0.790)
2
=0.6241
s2
2
= (0.940)
2
=0.8836
Null hypothesis is as follows
H0=there is no significant difference between the variance of variances of GPA for students who
completed the course or who didn’t. i.e.
σ1
2
≠ σ2
2
Against the alternative,
H1= σ1
2
=σ2
2

Test statistics is given by
  1 2
2 2
1 1
; 1 12 2
2 2
n n
s
F F
s


  

Under H0 test statistics is   1 2
2
1
; 1 12
2
n n
s
F F
s
  

Where larger value is placed in numerator. In our case F is,
.05;16,60
0.8836
1.4157
0.6241
F F  =calculated value
So F tabulated value for .05 level of significance at 16, 60 degrees of freedom is =1.82.
Since our calculated value is less than tabulated value, it is not significant. Hence we accept the
null hypothesis.
P value for given problem is=0.1655.
p-value >α so we fail to reject null hypothesis. And accept it. Result is same in both the cases.
Answer 4)
We need to apply chi-square test for goodness of fit in this problem.
Null hypo: county type not independent of days of week in receiving call
Alternate hypo: county type independent of days of week in receiving call
Level of significance=.005
Days of the week urban rural sum
sun 62 6 68
mon 47 10 57
tue 48 18 66
wed 51 17 68
thu 61 11 72
Chi –square table
county Days Observed
frequency
Expected
frequency
urban Sun 62
55.43949
Mon 47
46.47134
Tue 48
53.80892
Wed 51
55.43949
Thu 61
58.70064
Fri 74
70.92994
Sat 41
43.21019
rural Sun 6
12.56051
Mon 10
10.52866
Tue 18
12.19108
Wed 17
12.56051
Thu 11
13.29936
Fri 13
16.07006
Sat 12
9.789809
Total 471
E(62)=68*384/471=55.43949 in the same way other observed frequencies are found.
fri 74 13 87
Sat 41 12 53
total 384 87 471
Answer 5) this is large sample problem testing of difference of proportion.
Here attribute of interest is ‘drug- resistance’ and sample size for first sample is n1=284 and no
of members having the characteristic of interest is r1=18 ,so we get sample proportion for first
sample as p1=18/284=0.06338.
Same with other sample n2=536, r2=10,...
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