problems

1. A major health care agency provides one of the most
comprehensive emergency medical services.
Operating in a multiple hospital system with approximately 20
mobile medical units, the service goal is to respond to medical
emergencies with a mean time of 12 minutes or less.
The EMS director wants to perform a hypothesis test, with a
0.05 level of significance, to determine whether the service
goal of 12 minutes or less is being achieved.
The response times for a random sample of 40 medical
emergencies were tabulated. The sample mean is 13.25
minutes. The population standard deviation is believed
to be 3.2 minutes.

2. The production line for Glow toothpaste is designed to fill
tubes with a mean weight of 6 oz. Periodically, a sample of 30
tubes will be selected in order to check the filling process.
Quality assurance procedures call for the continuation of the
filling process if the sample results are consistent with the
assumption that the mean filling weight for the population of
toothpaste tubes is 6 oz.; otherwise the process will be
adjusted.
Assume that a sample of 30 toothpaste tubes provides a
sample mean of 6.1 oz. The population standard deviation is
believed to be 0.2 oz.
Perform a hypothesis test, at the 0.03 level of significance, to
help determine whether the filling process should continue
operating or be stopped and corrected.

3. An investment services company claims that the average
annual return on stocks within a certain industry is 11.5%.
An investor wants to test whether this claim is true and
collects a random sample of 30 stocks in the industry of
interest. He finds that the sample average annual return is
10.8% and that the sample standard deviation is 3.4%.
Does the investor have enough evidence to reject the
investment company’s claim? (Use a = 0.05)

4. The theory of finance allows for the computation of
“excess” returns, either above or below the current stock
market average. An analyst wants to determine whether
stocks in a certain industry group earn either above or
below the market average at a certain time period. The null
hypothesis is that there is no excess returns, on the average,
in the industry in question. “No average excess returns”
means that the population excess return for the industry is
zero. A random sample of 24 stocks in the industry reveals
a sample average excess return of 0.12 and sample standard
deviation of 0.2. State the null and alternate hypotheses,
and carry out a test at a = 0.05.

5. The marketing manager of a company wishes to estimate the
average sales per dealer. He is prepared to accept the error of
the estimate to be more than 5% of the actual average with
probability 0.05. While his boss insists that he would not
accept an error above 1% of the actual average with the same
confidence level. To satisfy his boss, by how many times
should he increase/decrease the sample size?

6. For a Christmas and New Year’s week, the National Safety
Council estimated that 500 people would be killed and
25,000 injured on the nation’s roads. The NSC claimed that
50% of the accidents would be caused by drunk driving.
A sample of 120 accidents showed that 67 were caused by
drunk driving. Use these data to test the NSC’s claim with
a = 0.05.

7. A city has a total population of one crore. It is estimated that
90% of the city dwellers use only toothpaste of one brand or
the other, while the remaining 10% use both tooth powder or
other things for tooth care. Several competing brands of
toothpaste are available in the city. In order to estimate the
market share of a particular brand X, i.e., the proportion of
people who use brand X amongst all toothpaste users
(focusing only on toothpaste users), what approximate size
of sample should be selected so that the probability that the
absolute error in the sample estimate lies within 5% of the true
market share is 0.95? An initial pilot survey of 100 toothpaste
users suggests that the market share of brand X is 25%.

8. The proportion of families buying milk from company A in a
certain city is believed to be P = 0.6. If a random sample of 100
families shows that 30 or less buy milk from company A, we
shall reject the hypothesis that H0: P = 0.6 in favor of the
alternative Ha: P < 0.6. Find the probability of committing type I
error. Evaluate the probability of committing type II error for
alternatives P = 0.3 and P = 0.5.

9. The government claims that about 25% of the rural
population in the state are below the poverty line. In order to
counter the government’s claim the main opposition party
conducts a quick survey of 1000 rural people and finds that
275 of them are below the poverty line. Can the opposition
conclude at 5% level of significance that the government’s
claim is false? What is the probability that the above survey
accepts the government’s claim while in reality 30% of the
rural population are below the poverty line?

10. A particular item is delivered in lots of 10000 pieces by a
company to the retailers. The retailers have agreed to accept a
lot with a maximum of 5% defective items. In case the
percenatge defective (P) in the lot is more than 5%, the retailer
sends back the entire lot to the company. To test whether a lot
conforms to specification, the company has decided to use a
sampling inspection plan as follows. A random sample of
items is drawn from the lot and the number defectives in the
sample is counted. If the number of defectives in the sample
exceed some pre-assigned number the lot is not sent to any
retailer. Determine the (approximate) number of items to be
inspected as well as the maximum number of defectives in the
sample up to which a lot remains acceptable so that the
probability of type I error is 0.05, and the probability of type
II error when P = 0.10 is 0.05.

11. The Controller of Examinations of a University considers that
the variation of marks in any subject should neither be too high
nor should it be too low. She has decided that the standard
deviation of marks in any subject should be around 10. To
verify if there is any violation of this norm, she has decided to
carry out a test on the basis of a random sample of 20 answer
books. For one subject, the sample mean and the standard
deviation are 60 and 7 respectively. What is you conclusion
regarding the stipulation of the Controller for this subject?

12. A manufacturer of doorknobs has a production process that is
designed to provide a doorknob with a target diameter of 2.5
inches. In the past, the S.D. of the diameter has been 0.035
inches. In an effort to reduce the variance in the process, various
studies have taken place that have resulted in a redesigned
process. Now it is intended to test if the new redesigned process
has achieved the desired objective. It is decided to take a sample
of 25 doorknobs from the new process. If the sample S.D. lies
below a limit such that only 5% of the time the test gives a
wrong decision when actually there is no reduction in variance,
it will be concluded that there is no reduction in variance. If the
sample produces an S.D. of 0.025 inch, compute the critical
limit. What is your conclusion?

13. Bhavana, an employee of Karuna Lights and Tubes, a manufacturer of electric bulbs,
is trying to estimate the average lifetime of bulbs produced on a particular day.
Based on the previous production, she had estimated the standard deviation (s) of
the lifetime. After taking a sample of 100, she had calculated a two-sided 95%
confidence interval whose lower limit turned out to be 830.5 hours. When the
management complained that the width of the confidence interval is too large, she
simply reduced the confidence level to 0.90 (still two-sided) and came up with a
lower limit of the confidence interval at 836.8 hours. The management is willing to
accept an estimate of the mean that is within 19.6 above or below the point estimate
but the chances of this estimate being incorrect must be only 1 in 20.
1. What was the original two sided 95% confidence interval of the lifetime?
2. What was the revised confidence interval with a confidence level of 0.90?
3. What should be the sample size to satisfy the requirements of the management?
4. When Bhavana selected a new sample with size as determined in (3) above, the
sample mean turned out to be 30 hours more than the previous sample mean.
Calculate the two-sided confidence interval as desired by the management, based on
the new sample.

14. Rahul, a young IT professional, is considering the possibility of
establishing an Educational Portal for prospective MBA
aspirants and wishes to conduct a survey. Based upon the cost of
setting up such a portal and the profits that may be generated, he
has arrived at the following conclusion: if there is evidence that
the average revenue per candidate is more than Rs.2500, then the
portal will be established; otherwise, the portal will not be
established. Based on past experience of several other such
portals, the standard deviation of revenue is estimated to be
Rs.500. Rahul wants to be 99% certain of not committing an
error of establishing the portal when the actual average revenue
is at most Rs.2500 per candidate. If Rahul wishes to have a 2.5%
chance of not establishing the portal when the actual average
revenue is Rs.3000, what sample size should be selected?

15. Management of the Modern Steel Co. wishes to determine if
there is any difference in performance between the day shift
workers and the evening shift workers. It is decided to take a
sample of 100 workers from each shift and compare their average
performance. If the difference in the average performance lies
within the limits such that only 10% of the time the test gives a
wrong decision when actually there is no difference in the
performance, it will be concluded that there is no difference in
performance. If for the day-shift workers, the sample average is
74.3 parts per hour with a sample S.D. of 16 parts per hour and
for the evening-shift workers, sample average and sample S.D.
are 69.7 and 18 parts per hour respectively, compute the critical
limits. What is your conclusion?

16. The marketing manager of a company carries out a survey by
taking a simple random sample from the user population of a
particular product, which has several competing brands in the
market, to estimate the market share of his company’s brand.
He finds that out of 1000 selected in the sample only 247 use
his company’s brand. Not satisfied with his company’s
performance he decides, after discussions with his colleagues,
to launch a promotion campaign for a month. After three
months, he carries out a second survey in which he finds that
512 out of 1500 users of the product use his company’s brand.
On the basis of the survey data can the marketing manager
conclude that the promotion campaign has yielded the desired
result?

17. A company’s market share is very sensitive to both its level of
advertising and the levels of its competitors’ advertising. A firm
known to have a 56% market share wants to test whether this
value is still valid in view of recent advertising campaigns of its
competitors and its own increased level of advertising. A random
sample of 500 consumers reveals that 298 use the company’s
product. Is there evidence to conclude that the company’s market
share is no longer 56%, at the 0.01 level of significance? What
is the probability that you conclude that there is no change in
market share when actually the market share has gone up to
60%?

18. A manufacturer of golf balls claims that the company
controls the weights of the golf balls accurately so that the
variance of the weights is not more than 1 mg2. A random
sample of 31 balls yields a sample variance of 1.62 mg2. Is
that sufficient evidence to reject the claim at an a of 5%?
What is the critical limit for the sample variance?

19. Airbus Industries, the European maker of the A320 medium-range
jet capable of carrying 150 passengers, is currently trying
to expand its market world-wide. At one point, Airbus managers
wanted to test whether their potential market in the United
States, measured by the proportion of airline industry executives
who would prefer the A320, is greater than the company’s
potential market for the A320 in Europe (measured by the same
indicator). A random sample of 120 top executives of US
airlines looking for new aircraft were given a demonstration of
the plane, and 34 indicated that they would prefer the model to
other new planes on the market. A random sample of 200
European airlines executives was also given a demonstration of
the plane, and 41 indicated that they would be interested in the
A320. Test the hypothesis that more US airlines executives
prefer the A320 than their European counterparts.

20. A banker with Continental India National Bank and Trust
Company wants to test which method of raising cash for
companies – borrowing from public sources or borrowing from
private sources – results in higher average amounts raised by a
company. The banker collects a random sample of 12 firms that
borrowed only from public sources and finds that the average
amount borrowed by a company per source is Rs.125 crores
and the standard deviation is Rs.34 crores. Another sample of
18 firms that borrowed only from private sources gives a
sample average per source of Rs.210 crores and a sample
standard deviation of Rs.50 crores. Do you believe that private
sources lend, on the average, twice the public sources?

21. One of the problems that insider trading supposedly causes is unnaturally
high stock price volatility. When insiders rush to buy a stock they believe
there will be increase in price, the buying pressure causes the stock price to
rise faster than under usual conditions. Then, when insiders dump their
buildings to realize quick gains, the stock price dips fast. Price volatility
can be measured as the variance of prices.
An economist wants to study the effect of the insider trading scandal and
ensuing legislation on the volatility of the price of a certain stock. The
economist collects price data for the stock during the period before the
event (interception and prosecution of insider traders) and after the event.
(The assumption of random sampling may be somewhat problematic in
this case, but later we will deal with time-dependent observations more
effectively.) The economist wants to test whether the event has changed
the variance of prices of the stock. The 25 daily stock prices before the
event give s2
1 = 9.3, and the 24 stock prices after the event give s2
2 = 3.0.

22. A metropolitan transit authority wants to determine whether there is any need
for changes in the frequency of service over certain bus routes. The transit
authority needs to know whether the frequency of service should increase,
decrease, or remain the same. It is determined that if the average number of
miles traveled by bus over the routes in question by all residents of a given
area is about 5 or more per day, then no change will be necessary. If the
average number of miles traveled per person per day is less than 5, then
changes in service may be necessary. Based on past records of similar
routes, it is known that the standard deviation of number of miles traveled
per person per day is 1.5 miles. The authority wants to be 99.9% confident of
not committing the error of making changes when there is no need for any
change. The authority also wishes to be 99% sure of initiating changes in
services when the number of miles traveled per person per day is actually 4.
How many residents should be selected in the sample? What is the critical
limit for the sample average number of miles traveled per person per day?

23. For every lot of 100 computer chips a company produces, an
average of 1.4 are defective. Another company buys many lots at
a time, from which one lot is selected randomly and tested for
defects. If the tested lot contains more than three defects, the
buyer will reject all the lots sent in that batch. What is the
probability that buyer will accept the lots?

24. A large insurance company wants to estimate the difference between the
average amount of term life insurance purchased per family and the average
amount of whole life insurance purchased per family. To obtain an estimate,
one of the company’s actuaries randomly selects 27 families who have term
life insurance only and 29 families who have whole life policies only. Each
sample is taken from families in which the leading provider is younger than
45 years of age. Use a 95% confidence interval to estimate the difference in
means for these two groups. Also determine whether the variances of term
and whole life insurance amounts are the same at 5% level of significance.
Term Whole
Life
Sample
Mean
Rs. 75000 Rs. 45000
Sample SD Rs. 22000 Rs. 15500

25. In order to estimate the number of additional telephone lines it
would need, a company has collected the following data
relating to the inter-arrival time (i.e., the time between two
consecutive arrival) of 100 calls (with sample mean = 2.5).
Using a suitable procedure, test if the inter-arrival time
distribution can be assumed to be exponential.
Inter-arrival
Time
(min.)
Less
than
1 min.
1 min –
3 min
3 min –
5 min
5 min –
7 min
7 min –
9 min
9 min –
11 min
Greater
than
11 min
Number
of calls
32 40 16 6 3 1 2

26. Following is the income distribution of a random sample of 200
households. Check if the income distribution can be assumed to
be normal.
Income
Group
No. of
Households
< 38 3
38 -40 2
40 - 42 7
42 - 44 13
44 - 46 14
46 - 48 24
48 - 50 29
50 - 52 27
52 - 54 32
54 - 56 20
56 - 58 15
58 - 60 10
60 - 62 3
62 - 64 0
> 64 1
200

27. A study of educational levels of voters and their political party
affiliations yielded the following results.
Party Affiliation
Educational
Level
Congress BJP Others
Below high
School
40 20 10
High school 30 35 15
College Degree 30 45 25
Determine whether party affiliation is independent of the
educational levels of the voters at 1% significance level.

28. A milk company has four machines that fill jugs with milk.
The quality control manager is interested in determining
whether the average fill for these machines is the same. The
following data represent random samples of fill measures (in
liters) for 19 jugs filled by the different machines. Use 5%
significance level.
Machine 1 Machine 2 Machine 3 Machine 4
4.05 3.99 3.97 4.00
4.01 4.02 3.98 4.02
4.02 4.01 3.97 3.99
4.04 3.99 3.95 4.01
4.00 4.00
4.00

29. Sample
Means
Samples
1 2 1.5
1 3 2
1 4 2.5
1 5 3
1 6 3.5
2 3 2.5
2 4 3
2 5 3.5
2 6 4
3 4 3.5
3 5 4
3 6 4.5
4 5 4.5
4 6 5
5 6 5.5
Mean 3.5
SD 1.080123
123
456
Population
Mean 3.5
SD 1.707825
1.080123
EE(( x )) == m
x s
s s
x n
N n
N
= -
-
( )
1

30. Histogram
4
3
2
1
0
1.5
2.5
3.5
4.5
5.5
Bin
Frequency
Frequency
Distribution of
Sample Means
1.5 1
2 1
2.5 2
3 2
3.5 3
4 2
4.5 2
5 1
5.5 1
15

31. Sample
Means
1 2 3 2
1 2 4 2.333333
1 2 5 2.666667
1 2 6 3
1 3 4 2.666667
1 3 5 3
1 3 6 3.333333
1 4 5 3.333333
1 4 6 3.666667
1 5 6 4
2 3 4 3
2 3 5 3.333333
2 3 6 3.666667
2 4 5 3.666667
2 4 6 4
2 5 6 4.333333
3 4 5 4
3 4 6 4.333333
3 5 6 4.666667
4 5 6 5
Mean 3.5
SD 0.763763
Samples
EE(( x )) == m
x s
123
456
Population
Mean 3.5
SD 1.707825
0.763763
s s
x n
N n
N
= -
-
( )
1

32. Distribution of
Sample Means
2 1
2.333333 1
2.666667 2
3 3
3.333333 3
3.666667 3
4 3
4.333333 2
4.666667 1
5 1
20
Histogram
0 1 2 3 4
2
2.666666667
3.333333333
4
4.666666667
More
Bin
Frequency
Frequency

33. Form of the Sampling Distribution of x
IIff wwee uussee aa llaarrggee ((nn >> 3300)) ssiimmppllee rraannddoomm ssaammppllee,, tthhee
cceennttrraall lliimmiitt tthheeoorreemm eennaabblleess uuss ttoo ccoonncclluuddee tthhaatt tthhee
ssaammpplliinngg ddiissttrriibbuuttiioonn ooff x
ccaann bbee aapppprrooxxiimmaatteedd bbyy
aa nnoorrmmaall ddiissttrriibbuuttiioonn..
WWhheenn tthhee ssiimmppllee rraannddoomm ssaammppllee iiss ssmmaallll ((nn << 3300)),,
tthhee ssaammpplliinngg ddiissttrriibbuuttiioonn ooff x
ccaann bbee ccoonnssiiddeerreedd
nnoorrmmaall oonnllyy iiff wwee aassssuummee tthhee ppooppuullaattiioonn hhaass aa
nnoorrmmaall ddiissttrriibbuuttiioonn..

37. Relationship Between the Sample Size
and the Sampling Distribution ofx
WWiitthh nn == 110000,,
E(x) = 990
x
WWiitthh nn == 3300,,
14.6 x s =
8 x s =

38. Type I and Type II Errors
PPooppuullaattiioonn CCoonnddiittiioonn
CCoorrrreecctt
DDeecciissiioonn TTyyppee IIII EErrrroorr
CCoorrrreecctt
AAcccceepptt HH00
((CCoonncclluuddee m << 1122))
TTyyppee II EErrrroorr DDeecciissiioonn RReejjeecctt HH00
((CCoonncclluuddee m >> 1122))
HH00 TTrruuee
((m << 1122))
HH00 FFaallssee
CCoonncclluussiioonn ((m >> 1122))

39. Lower-Tailed Test About a Population Mean:
Reject H0
a = .10
Sampling
distribution
of z x
= n -m
0
/
Do Not Reject H0
-z 0 a = -1.28
zz
s
s Known
• Critical Value Approach

40. Upper-Tailed Test About a Population Mean:
a = .05
0 za = 1.645
Reject H0
0
/
Do Not Reject H0
zz
Sampling
distribution
of z x
= -m
s
n s Known
• CCrriittiiccaall VVaalluuee AApppprrooaacchh

41. Two-Tailed Tests About a Population Mean:
 Critical Value Approach
0
/
Do Not Reject H Reject H0 0
a/2 = .015
0 2.17
z
Reject H0
-2.17
Sampling
distribution
of z x
= -m
s
n s Known
a/2 = .015

42. One-Tailed Tests About a Population Mean:
• pp ––VVaalluuee AApppprrooaacchh
p-value
= .0068
0 za =
1.645
a = .05
z
z =
2.47
s Known
Sampling
distribution
of z x
= n -m
0
/
s

43. Two-Tailed Tests About a Population Mean:
s Known
a/2 =
.015
z = -2.74 0
z = 2.74
za/2 = 2.17
z
 p-Value Approach
a/2 =
.015
-za/2 = -2.17
1/2
p -value
= .0031
1/2
p -value
= .0031

44. Probability of a Type II Error
m0
a
ma
x
x
Sampling
distribution
of x
when
His true
0 and m = m0
Sampling
distribution
of x
when
His false
0 and m> ma 0
Reject H0
b
cc
cc
HH00:: m < m0
HHaa:: m > m0
s = s x n NNoottee::

45. Probability of a Type II Error
a = 0.05
k
Reject H0
Do Not Reject H0
b
xx__bbaarr
0 m
a m

46. Probability of a Type II Error
k
a = 0.05
Reject H0
Do Not Reject H0
b
xx__bbaarr
0 m
a m

47. Probability of a Type II Error
k
a = 0.05
Reject H0
Do Not Reject H0
b
xx__bbaarr
0 m
a m

48. Calculating the Probability
of a Type II Error
PPrroobbaabbiilliittiieess tthhaatt tthhee ssaammppllee mmeeaann wwiillll bbee
iinn tthhee aacccceeppttaannccee rreeggiioonn::
12.8323
3.2/ 40
z = -m
VVaalluueess ooff m b 11--b
1144..00 --22..3311 ..00110044 ..99889966
1133..66 --11..5522 ..00664433 ..99335577
1133..22 --00..7733 ..22332277 ..77667733
1122..88332233 00..0000 ..55000000 ..55000000
1122..88 00..0066 ..55223399 ..44776611
1122..44 00..8855 ..88002233 ..11997777
12.0001 1.645 ..99550000 ..00550000

49. Power Curve
1.00
0.90
Rejecting Null Hypothesis m
Correctly
0.80
HFalse
0.70
0 0.60
of 0.50
Probability 0.40
0.30
0.20
0.10
0.00
11.5 12.0 12.5 13.0 13.5 14.0 14.5