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Kuwait University
College of Business Administration
Department of Quantitative Methods and information System
Tutorial
Stat 220
Chapter 7: sampling and sampling distribution
Chapter 8: confidence interval
Chapter 9: hypothesis testing
Chapter 10: statistical inference based on two samples
Chapter 11: ANOVA
Chapter 12: chi-square test for independence
Chapter 13: simple linear regression
Chapter 14: multiple regression and model building
Done by:
T.A Dalal Al-Odah
T.A Narjes Akbar
T.A Dalal AL-Banwan
Supervised by Dr.Mohammed Qadry Grraph
summer
2014/2015
1
Chapter 7
I.
̅
Sampling distribution of sample mean 𝒙
Exactly normal
Approximately normal
 If the population data follow
Normal (𝜇, 𝜎)
Then
 If the population data follow
any distribution (not normal)
and the sample size is large
(𝑛 ≥ 30)
Then
 The sampling distribution
𝜎
𝑥̅ ~𝑁 (𝜇𝑥̅ = 𝜇, 𝜎𝑥̅ = )
√𝑛
 The sampling distribution
𝜎
𝑥̅ ~𝑁 (𝜇𝑥̅ = 𝜇, 𝜎𝑥̅ = )
√𝑛
(by central limit theorem CLT)
Where:
𝑥̅ : Sample mean
𝑍=
𝑥̅ − 𝜇
𝜎
√𝑛
𝜇: Population mean
𝜎: Population standard deviation
𝜎𝑥̅ : Standard deviation of the sample mean (standard error)
𝜇𝑥̅ : mean of the sample mean
Note: mean = average = rate = expected
2
EXERCISES
1. The amounts of electric bills for all households in a city have a skewed probability
distribution with a mean of $80 and a standard deviation of $25. Let 𝑥̅ be the mean
amount of electric bills for a random sample of 75 households selected from this city.
Find:
a. The mean of the sampling distribution of 𝑥̅
b. The variance of the sampling distribution of 𝑥̅
c. The standard deviation (error) of the sampling distribution of 𝑥̅
d. What is the sampling distribution of the sample mean 𝑥̅
e. Find the probability that the mean amount of electric bills for a random sample of
75 households selected from this city will be
i. Between $72 and $77
ii. Within $6 of the population mean
iii. More than the population mean by at least $5
iv. Less than the population mean by at least $2
v. Either less than $72 or more than $77
3
2. The print on the package of 100-watt General Electric soft-white light bulbs claims that
these bulbs have an average of 750 hours. Assume that the lives of such bulbs have a
normal distribution with a mean of 750 hours and a standard deviation of 55 hours. Let 𝑥̅
be the mean life of a random sample of 25 such bulbs.
a. Find the mean and standard deviation of 𝑥̅ and describe its sampling distribution.
b. Find the probability that the mean life of a random sample of 25 such bulbs will
be within 15 hours of the population mean.
c. Find the fraction that 𝑥̅ will be less than the population mean by 20 hours or more.
d. Find the percentage that 𝑥̅ will be more than the population mean by at least 20
hours.
e. Find the probability that 𝑥̅ will be within 1.5 standard deviation (error) of the
population mean.
REVIEW
1. A quality control inspector periodically checks a production process. This inspector selects simple
random samples of 30 finished products and computes the sample mean product weights ̅𝑥. Test
results over long period of time show that 2.5% of the 𝑥̅ values are over 2.1 kg and 2.5% are less than
1.9 kg.
a. What are the mean and standard deviation for the population of products produced with this
process? (𝝁 = 𝟐, 𝝈 = 𝟎. 𝟐𝟕𝟗𝟓).
b. Find the probability that 𝑥̅ will be within one standard deviation (error) of the population
̅ ≤ 𝝁 + 𝝈𝒙̅ ) = 𝒑(−𝝈𝒙̅ ≤ 𝒙
̅ − 𝝁 ≤ 𝝈𝒙̅ ) = 𝒑(−𝟏 ≤ 𝒛 ≤ 𝟏) = 𝟎. 𝟔𝟖𝟐).
mean?(𝒑(𝝁 − 𝝈𝒙̅ ≤ 𝒙
2. A machine makes 3-inch-long nails. The probability distribution of the lengths of these nails is
normal with a mean of 3 inches and a standard deviation of 0.1 inch. The quality control inspector
takes a sample of 25 nails once a week and calculates the mean length of theses nails. If the mean of
this sample is either less than 2.95 inches or greater than 3.05 inches, the inspector concludes that the
machine needs an adjustment.
a. What is the mean, standard deviation (error) and sampling distribution of the sample
̅~𝑵(𝝁𝒙̅ = 𝟑, 𝝈𝒙̅ = 𝟎. 𝟎𝟐))
mean?(𝒙
b. What is the probability that based on a sample of 35 nails the inspector will conclude that the
̅ < 𝟐. 𝟗𝟓) + 𝒑(𝒙
̅ > 𝟑. 𝟎𝟓) =. 𝟎𝟎𝟑)
machine needs an adjustment?(𝒑(𝒙
4
II.
̂
Sampling distribution of the sample proportion 𝑷
Approximately normal
 If
𝑛𝑃 ≥ 5
and
𝑛(1 − 𝑃) ≥ 5
Then
 The sampling distribution
𝑃̂~𝑁(𝜇𝑃̂ = 𝑃, 𝜎𝑃̂ = √
𝑃(1 − 𝑃)
)
𝑛
(by central limit theorem CLT)
Where:
𝑍=
𝑃̂ − 𝑃
√𝑃(1 − 𝑃)
𝑛
𝑃: Population proportion
𝑃̂: Sample proportion
𝜇𝑃̂ : Mean of the sample proportion
𝜎𝑃̂ : Standard deviation of the sample proportion (standard error)
5
EXERCISES
1. A corporation makes auto batteries. The company claims that 80% of its batteries are
good for 70 months or longer. Assume that this claim is true. Let 𝑃̂ be the proportion in a
sample of 100 batteries that are good for 70 months or longer.
a. What is the mean, standard deviation (error), and sampling distribution of the
sample proportion?
b. The probability that the proportion is less than 0.09?
c. The probability that this 𝑃̂ is within 0.05 of the population proportion?
d. The probability that this 𝑃̂is not within 0.05 of the population proportion?
e. What is the probability that 𝑃̂ is less than the population proportion by 0.06 or
more?
REVIEW
1. Suppose that among the undergraduate students at a very large university 5.9% are international
students and 57.8% are female.
a. If 28 students are randomly sampled, what is the probability that fewer than 14 are
𝒙
𝟏𝟒
̂𝒇 < 𝟎. 𝟓) = 𝑷(𝒛 < −𝟎. 𝟖𝟒) = 𝟎. 𝟐𝟎𝟎𝟓).
female?(𝑷(𝒙𝒇 < 𝟏𝟒) = 𝑷 ( 𝒏𝒇 < 𝟐𝟖) = 𝑷(𝒑
b. If 10 students are randomly sampled, what is the probability that more than 10% are
̂ 𝑰 > 𝟎. 𝟏) = 𝑷(𝒛 > 𝟎. 𝟓𝟓) = 𝟎. 𝟐𝟗𝟏𝟐).
international students?(𝑷(𝑷
2. A machine that is used to make CDs is known to produce 6% defective CDs. The quality control
inspector selects a sample of 100 CDs every week and inspects them for being good or defective.
If 8% or more of the CDs in the sample are defective, the process is stopped and the machine will
be adjusted. What is the probability that based on a sample of 100 CDs the process will be
̂ > 𝟎. 𝟎𝟖) = 𝑷(𝒛 > 𝟎. 𝟖𝟑) = 𝟎. 𝟐𝟎𝟑𝟑).
stopped to adjust the machine?(𝑷(𝒑
6
Chapter 8
I.
Population mean (𝝁)
̅
̂) = 𝒙
 point estimate of 𝜇 (𝝁

Confidence interval (C.I.) of 𝝁
𝜎 Unknown
𝜎 known
𝑛 ≥ 30
𝑛 < 30
𝑥̅ ± 𝑡𝛼⁄2,𝑛−1
𝑠
𝑥̅ ± 𝑍𝛼⁄2
√𝑛
𝑥̅ ± 𝑍𝛼⁄2
𝜎
√𝑛
𝑠
√𝑛
Margin of Error
(Max Error/ Error)
 To fine the sample size
𝑍𝛼⁄2 𝑠𝑡𝑑. 𝑑𝑒𝑣 2
𝑛=(
)
𝐸
 If we have C.I. ( L , U ) then
o Sample Mean
o Marginal error
𝐿+𝑈
2
𝑈−𝐿
2
=
𝑤𝑖𝑑𝑡ℎ (𝑙𝑒𝑛𝑔𝑡ℎ)
2
7
Exercises:
1. A random sample of 16 mid-sized cars, which were tested for fuel consumption, gave
a mean of 26.4 miles/gallon with standard deviation of 2.3 miles/gallon.
a. Find a 95% confidence interval for the average fuel consumption of a midsized car?
b. What assumption(s) are necessary for your answer in (a) to be valid?
c. Find the error of such interval?
d. If we choose a sample of size 100 mid-sized cars, then repeat part (a)?
e. What sample size would be required to reduce the margin of error by 50%?
2. An economist wants to find a 90% confidence interval for the mean sale price of
houses in a state. How large a sample should he or she select so that the estimate is
within $3500 of the population mean? Assume that the standard deviation for the sale
prices of all houses in this state is $31500?
8
3. IQ tests are designed to yield results that are approximately normally distributed.
Researchers think that the population standard deviation is 15. A reporter is interested
in estimating the average IQ of employees in a large high-tech firm in California. She
gathers the IQ information on 22 employees of this firm and records the sample mean
IQ as 106.
a. Compute 90% confidence intervals of the average IQ in this firm.
b. If the C.I is (97.77, 114.23) find the confidence level
4. In analyzing the operating cost for a huge fleet of delivery trucks, a manager takes a
sample of 25 cars and calculated the sample mean and variance of the operating cost.
Under the assumption that the operating cost has a normal distribution, he found that
the 95% confidence interval for the mean operating cost is between 253 and 300 K.D.
a. Find the maximum error of estimate (error bound) for such interval?
b. Find the sample mean and variance?
c. The manager said he is 95% confidence that the sample mean lies within such
interval, do you agree? Why?
d. Construct a 90% confidence interval for the true mean? Find the error of such
interval?
9
Review:
1. To measure the time taken to manufacture a device, a random sample was chosen. The
following is the assembly time (the time taken to fix each device in minutes ) for the
sample:
8
10
12
15
17
If the sample information is used to estimate the population mean of the assembly time
then:
a. Give a point estimate and a 99% confidence interval for the population mean?
∑
𝒔
State your necessary assumptions we need?(𝝁̂ = 𝒙̅ = 𝒏𝒙 = 𝟏𝟐. 𝟒, 𝒙̅ ± 𝒕𝜶⁄𝟐,𝒏−𝟏 √𝒏
,
𝒂𝒔𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏 𝒊𝒔 𝒏𝒐𝒓𝒎𝒂𝒍 𝒑𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 ).
b. If the population standard deviation is known to be 3, how large is the sample size
needed to estimate the mean assembly time with 0.99 confidence, and error
margin of one minute?(𝒏 = (
𝒁𝜶⁄ 𝝈 𝟐
𝟐
𝑬
) =(
𝟐.𝟓𝟕𝟔×𝟑 𝟐
𝟏
) = 𝟓𝟗. 𝟕𝟐 ≅ 𝟔𝟎)
2. Determine the margin of error for a confidence interval estimate for the population mean
of a normal distribution given the following information:
Confidence level=0.98, n=13, S=15.68 (𝑴. 𝑬 = 𝟏𝟏. 𝟔𝟔)
10
II.
Population proportion
 Point estimate of P= 𝑃̂
 Confidence interval of P
𝑃̂ (1 − 𝑃̂)
𝑃̂ ± 𝑍𝛼⁄2 √
𝑛
Margin of Error
(Max Error/ Error)
 To find the sample size
𝑛=(
𝑍𝛼⁄ 2 𝑝̂𝑞̂
2
𝐸2
)
Where 𝑞̂ = 1 − 𝑝̂
11
Exercises:
1. It is said that happy and healthy workers are efficient and productive. A company that
manufactures exercising machines wanted to know the percentage of large companies
that provide on-site health club facilities. A sample of 240 such companies showed that
96 of them provide such facilities on site.
a. What is the point estimate of the percentage of all such companies that provide
such facilities on site? What is the margin of error associated with this point
estimate?
b. Construct a 97% C.I for the percentage of all such companies that provide such
facilities on site.
2. A consumer agency wants to estimate the proportion of all drivers who wear seat belts
while driving. Assume that a preliminary study has shown that 76% of drivers wear seat
belts while driving. How large should the sample size be so that the 99% C.I for the
population proportion has a maximum error of 0.03?
3. A college registrar has received complaints about the online registration procedure at her
college. She wants to estimate the proportion of all students at this college who are
dissatisfied the online registration procedure. What is the most conservative estimate of
the sample size that would limit the maximum error to be within 0.05 of the population
proportion for 90% C.I?
12
Review
1. A researcher wanted to know the percentage of judges who are in favor of the death
penalty. He took a random sample of 15 judges and asked them whether or not they favor
of the death penalty. The responses of these judges are given here
Yes
No
Yes
Yes
No
No
No
Yes
Yes
No
Yes
Yes
Yes
No
Yes
a. What is the point estimate of the population proportion? What is the margin of
̂
̂
̂ = 𝒙 = 𝟎. 𝟔, 𝑴. 𝑬 = 𝒁𝜶⁄ √𝑷(𝟏−𝑷) = 𝟎. 𝟐𝟒𝟕𝟗)
error associated with this point estimate?(𝑷
𝟐
𝒏
𝒏
b. Make a 95% C.I for the percentage of all judges who are in favor of the death
penalty.(𝟎. 𝟔 ± 𝟎. 𝟐𝟒𝟕𝟗𝟐)
2.
a. How large a sample should be selected so that the maximum error of estimate for
99% C.I for the population proportion is 0.035? When the value of the sample
proportion obtained from a preliminary sample is 0.29?(𝟎. 𝟐𝟗±. 𝟎𝟑𝟓)
b. Find the most conservation sample size that will produce the maximum error for a
99% C.I for p equal to 0.035(𝑯𝒊𝒏𝒕: 𝑖𝑓 𝑝̂ 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑡𝑎𝑘𝑒 𝑝̂ = 0.5, 𝒏 = 𝟏𝟑𝟓𝟒. 𝟐𝟓 ≅ 𝟏𝟑𝟓𝟓)
13
Chapter 9
I. Testing hypothesis for 𝝁
1. State null hypothesis (𝐻0 ) and alternative hypothesis (𝐻1 )
𝐻0 : 𝜇 =
𝐻1 : 𝜇 ≠
<
>
vs
≥
≤
T.S for 𝜇
2. Calculate the test statistic (T.S)
𝜎 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝑛 < 30
𝑇𝐶 =
𝜎 𝑘𝑛𝑜𝑤𝑛
𝑍𝐶 =
𝑛 ≥ 30
𝑥̅ − 𝜇
𝑠
√𝑛
𝑍𝐶 =
𝑥̅ − 𝜇
𝑠
√𝑛
𝑥̅ − 𝜇
𝜎
√𝑛
3. Calculate P-value
3. Determine the rejection region (R.R)
p-value
+
p-value
+
−𝑍𝛼⁄
2
𝑍𝛼⁄
2,𝑛−1
𝑡𝛼⁄2,𝑛−1
−𝑡𝛼⁄
2
Two-Tailed
−𝑍𝛼
−𝑡𝛼,𝑛−1
Left-Tailed
4. Conclusion: we reject 𝐻0 if T.S lies in the R.R
p-value
𝑍𝛼
𝑡𝛼,𝑛−1
Right-Tailed
−𝑍𝐶
𝑍𝐶
Two-Tailed
−𝑍𝐶
𝑍𝐶
Left-Tailed
4. Conclusion: we reject 𝐻0 if
𝑝𝑣𝑎𝑙𝑢𝑒 < 𝛼
Right-Tailed
14
 Type І error:
o What is the type І error?
𝑅𝑒𝑗𝑒𝑐𝑡 𝐻0 / 𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒
o What is the probability of type І error?
P (type І error) = α = 𝑃(𝑅𝑒𝑗𝑒𝑐𝑡 𝐻0 / 𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒)
o Note: 1 − 𝛼 = 𝑃(𝑑𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 /𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒)
 Type П error:
o What is the type П error?
𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 / 𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒
o What is the probability of type П error?
P (type П error) = β= 𝑃(𝐷𝑜 𝑛𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 / 𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒)
o Note: 1 − 𝛽 = 𝑃( 𝑅𝑒𝑗𝑒𝑐𝑡 𝐻0 /𝐻0 𝑖𝑠𝑓𝑎𝑙𝑠𝑒)
Power of
the test
The null hypotheses is
Your Decision
based on a random
sample
true
false
Reject
Type І error
Correct decision
Do not Reject
Correct decision
Type П error
15
Exercises:
1. According to survey by the national retail Association, the average amount that
households in the United States planned to spend on gifts, decorations, greeting cards,
and food during 2001 holiday season was $940. Suppose that a recent random sample of
324 households showed that they plan to spend an average of $1005 on such items during
this year’s holiday season with a standard deviation of $330.
a. Test at the 1% significance level whether the mean of such planned holiday
related expenditures for households for this year differs from $940.
1)
2)
3)
4)
b. Find 99% C.I of µ.
c. Use C.I from part (b) to test 𝐻0 : 𝜇 = 940
𝑣𝑠
𝐻1 : 𝜇 ≠ 940
16
2. A drug company is considering marketing a new local anesthetic. The effective time of
the anesthetic the drug company that is currently produced has a normal distribution with
an average of 7.4 minutes with standard deviation of 1.2 minutes. To market the new
anesthetic, the mean effective time should be less than 7.4 min. a sample of size 36
results in a sample mean of 7.1. a hypothesis test will be done to help make a decision.
a. State the null and the alternative hypotheses
b. Compute the test statistic
c. Compute the P-value of the test
d. What is your recommendation to the drug company using a level of significance of
0.01?
3. Insurance companies track life expectancy information to assist in determining the cost of
life insurance policies. Last year the average life expectancy of all policyholders was 77
years. A company wants to determine if their clients now have longer life expectancy on
average, so they randomly sample 20 of their recent paid policies. The sample has a mean
of 78.6 years and a standard deviation of 4.48 years.
a. Write the null and alternative hypotheses
b. What is the value of the test statistic?
c. State your conclusion using 𝛼=0.05
d. Considering the result of the test, which type of errors in hypothesis testing could
you have made?
e. State your assumptions
17
II.
Testing hypothesis for P
1. State null hypothesis (𝐻0 ) and alternative hypothesis (𝐻1 )
𝐻0 : 𝑃 =
vs
𝐻1 : 𝑃 ≠
<
≥
>
≤
2. Calculate the test statistic (T.S)
T.S for 𝑃
𝑍𝐶 =
𝑃̂ − 𝑃
√𝑃(1 − 𝑃)
𝑛
3. Calculate P-value
3. Determine the rejection region (R.R)
p-value
+
p-value
+
−𝑍𝛼⁄2
𝑍𝛼⁄2
Two-Tailed
−𝑍𝛼
Left-Tailed
𝑍𝛼
Right-Tailed
−𝑍𝐶
𝑍𝐶
Two-Tailed
4. Conclusion: we reject 𝐻0 if T.S lies in the R.R
p-value
−𝑍𝐶
𝑍𝐶
Left-Tailed
4. Conclusion: we reject 𝐻0 if
𝑝𝑣𝑎𝑙𝑢𝑒 < 𝛼
Right-Tailed
18
Exercises:
1. A food company is planning to market a new type of frozen yogurt. However,
before marketing this yogurt, the company wants to find what percentage of the
people like it. The company’s management has decided that it will market this
yogurt only if at least 35% of the people like it. The company’s research
department selected a random sample of 400 persons and asked them to taste this
yogurt. Of these 400 persons, 112 said they like it.
a. Testing at the 2.5% significant level, can you conclude that the company
should market this yogurt?
1)
2)
3)
4)
b. What will your decision be in part (a) if the probability of making a type І
error is zero?
2. A study by consumer reports showed that 64percent of supermarket shoppers
believes supermarket brands to be as good as national name brands.
a. Formulate the hypotheses that can be used to determine whether the
percentage of supermarket shoppers who believe that supermarket brands to
be as good as national name brands is different from 64%.
b. If a sample of 100 shoppers showed that 52 stating that the supermarket
brand was as good as the national brand, what is the value of the test
statistic?
19
c. What is the p-value?
d. At 𝛼 =0.05, what is your conclusion? Justify your answer.
3. Suppose that in a sample of 1000 employees 23% said that losing their job is the
major reason of concern for them.
a. Find a 98% confidence interval for the percentage of employees who said
losing their job is the major reason of concern for them.
b. According to your confidence interval obtained in (a) do you believe that
percentage is different from 19% and why or why not?
Review:
1. The policy of a company is to deliver on time at least 90% of all the orders it
receives from its customers. The quality control inspector at the company usually
takes samples of orders delivered and checks if this policy is maintained. A sample
of 90 orders taken by this inspector showed that 75 of them were delivered on
time. At the 2% significance level, can you conclude that the company’s policy is
maintained? Use the p-value to conduct the test.(𝑯𝟏 : 𝑷 < 𝟎. 𝟗, 𝒁𝒄 = −𝟐. 𝟐𝟏, 𝒑 −
𝒗𝒂𝒍𝒖𝒆 = 𝟎. 𝟎𝟏𝟑𝟔, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 ).
20
one population variance 𝝈𝟐
III.
 Point estimate for
one population variance 𝜎 2
one population st. deviation 𝜎
𝑆2
𝑆
 Confidence interval for
One population variance 𝜎 2
One population St. Deviation 𝜎
(𝑛 − 1)𝑠 2
(𝑛 − 1)𝑠 2
2
<𝜎 < 2
𝑥𝛼2⁄ ,𝑛−1
𝑥1−𝛼⁄ ,𝑛−1
2
(𝑛 − 1)𝑠 2
(𝑛 − 1)𝑠 2
<𝜎<√ 2
√ 2
𝑥𝛼⁄ ,𝑛−1
𝑥1−𝛼⁄ ,𝑛−1
2
2
2
 Hypothesis test about one variance 𝜎 2
1. 𝐻0 : 𝜎2 =
𝐻1 : 𝜎2 ≠
vs
≥
<
≤
>
2. Calculate the test statistic (T.S)
𝑥𝑐2
(𝑛 − 1)𝑆 2
=
𝜎2
3. Determine the rejection region
00
2
𝑥1−
𝛼⁄
2,𝑛−1
𝑥𝛼2⁄
2,𝑛−1
2
𝑥1−𝛼,𝑛−1
𝑥𝛼2 ,𝑛−1
4. Conclusion: Reject 𝐻0 if T.S 𝑥𝐶2 lies in rejection region
Note: hypothesis test about one population st.dev(𝜎)is the same as hypothesis test about
one population variance(𝜎 2 ), but you need to convert the hypothesis from 𝜎 to𝜎 2 .
21
Exercises:
1. A professor claims that the variance of the lengths of his lectures is within 2
square minutes. A random sample of 23 of these lectures was timed, and the
variance of the lengths of these lectures was found to be 2.7 square minutes.
Assume that the lengths of all such lectures by the professor are approximately
normally distributed.
a) Find the point estimate of the population variance
b) Make the 98% confidence intervals for the variance and standard deviation
of the lengths of all lectures by the professor.
c) Test at the 1% significance level whether the variance of the lengths of all
such lectures by the professor exceeds 2 square minutes.
22
2. An assembly line produces units with a mean weight of 10 and a standard
deviation of 0.20. A new process supposedly will produce units with the same
mean and a smaller standard deviation. A sample of 20 units produced by the new
method has a sample standard deviation of 0.126. At a significance level of 10%
can we conclude that the new process has less variation than the old?
Review
1. Automotive part must be machined to close tolerances to be acceptable to
customers. Production specifications call for a maximum variance in the
lengths of the parts of 0.0004. Suppose the sample variance for 30 parts turns
out to be 𝑠 2 = 0.0005. Using α=0.05, test to see whether the population
variance specification is being violated.
(𝑯𝟏 : 𝝈𝟐 > 𝟎. 𝟎𝟎𝟎𝟒, 𝒙𝟐𝒄 = 𝟑𝟔. 𝟐𝟓, 𝒙𝟐𝒕𝒂𝒃𝒍𝒆 = 𝟒𝟐. 𝟓𝟓𝟕, 𝒅𝒐 𝒏𝒐𝒕 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
23
Chapter 10
I.
The difference between two population means (𝝁𝟏 − 𝝁𝟐 ) for independent samples
𝜎1 & 𝜎2 unknown
𝜎1 & 𝜎2 known
𝑛1 & 𝑛2 Small
𝑛1 & 𝑛2 Large
 Point estimate:
𝜎12
≠
𝜎12
𝜎22
=
𝜎22
 Point estimate:
̅𝑋1 − 𝑋̅2
(Homogenous)
 C.I:
 Point estimate:
 Point estimate:
̅𝑋1 − 𝑋̅2
 C.I:
𝑆2 𝑆2
(𝑋̅1 − 𝑋̅2 ) ± 𝑍𝛼⁄ √ 1 + 2
2 𝑛
𝑛2
1
 C.I:
𝑆12 𝑆22
+
𝑛1 𝑛2
(𝑋̅1 − 𝑋̅2 ) ± 𝑡𝛼⁄
∗√
2,𝑛
 Test statistic :
( ̅𝑋1 − 𝑋̅2 ) − 𝐷
𝑡𝑐 =
𝑆2 𝑆2
√ 1+ 2
𝑛1 𝑛2

̅𝑋1 − 𝑋̅2
2
𝑆2 𝑆2
( 1 + 2)
𝑛1 𝑛2
2
2
𝑆2
𝑆2
( 1)
( 2)
𝑛1
𝑛
+ 2
𝑛1 − 1 𝑛2 − 1
𝑡𝑐 ~𝑡(𝑛∗ )
1
1
+
𝑛1 𝑛2
. 𝑆𝑃. √
2,𝑛1 +𝑛2 −2
 Pooled estimate:
(𝑛1 − 1)𝑆12 + (𝑛2 − 1)𝑆22
𝑆𝑃2 =
𝑛1 + 𝑛2 − 2
 Test statistic :
( ̅𝑋1 − 𝑋̅2 ) − 𝐷
𝑡𝑐 =
1
1
𝑆𝑃√ +
𝑛1 𝑛2
Degree of freedom:
𝑛∗ =
(𝑋̅1 − 𝑋̅2 ) ± 𝑡𝛼⁄
 Test statistic :
( ̅𝑋1 − 𝑋̅2 ) − 𝐷
𝑍𝑐 =
𝑆2 𝑆2
√ 1+ 2
𝑛1 𝑛2
̅𝑋1 − 𝑋̅2
 C.I:
𝜎2 𝜎2
(𝑋̅1 − 𝑋̅2 ) ± 𝑍𝛼⁄ √ 1 + 2
2 𝑛
𝑛2
1
 Test statistic :
( ̅𝑋1 − 𝑋̅2 ) − 𝐷
𝑍𝑐 =
𝜎2 𝜎2
√ 1+ 2
𝑛1 𝑛2
𝑍𝑐 ~𝑁(0,1)
𝑍𝑐 ~𝑁(0,1)
𝑡𝑐 ~𝑡(𝑛1 +𝑛2 −2)
24
II.
1. 𝐻0 : 𝜎21 = 𝜎22
Hypothesis test about Homogeneity
(Equal population variances 𝜎12 = 𝜎22 )
𝐻1 : 𝜎21 ≠ 𝜎22
vs
2. Calculate the test statistic (T.S)
𝐹𝐶 =
2
𝑆𝑙𝑎𝑟𝑔𝑒
2
𝑆𝑠𝑚𝑎𝑙𝑙
3. Determine the critical value
𝐹𝛼⁄2,
𝑛−1, 𝑛−1
Numerator
Denominator
4. Conclusion:
Reject 𝐻0 if T.S (𝐹𝐶 ) lies under the rejection region (under shaded area).
25
Exercises:
1. El-Mraay Dairy company claims that its 8-ounces low-fat yogurt cups contain on the
average fewer calories than the 8-ounces low-fat yogurt cups produced by its
competitor El-Safy company. In order to check this claim a sample of 50 such cups
produced by El-Mraay showed that they contains on the average 144 calories per cup
with a standard deviation of 5.4 calories. A sample of 40 cups of El-Safy product
showed that they contain on the average 147 calories per cup with a standard
deviation 6.3 calories.
a. Make a 98% confidence interval for the difference between the mean number
of calories in the 8-ounces low fat yogurt cups produced by the two
companies
b. Find the standard and margin error of part (a).
c. Does your C.I obtained in part (a) support the hypothesis that the two means
are different, what is the probability of type І error in that case.
d. Test El-Mraay Dairy Company claim with α=0.05
26
2. Two brands (A and B) of tires are tested to compare their durability. The management
of company claims that brand A is durable than brand B. Twelves from each brand
are tested on a machine. The mileages (in hundreds of miles for each tire) have been
recorded giving the following information.
Mileages in hundreds
Brand A 157 139 188 143 172 144 191 128 177 160 175 162
Brand B 160 118 150 165 158 159 127 133 170 164 152 142
Brand A
Brand B
Mean
161.3
149.8
Standard deviation
20.01
16.45
Assuming that two population are normally distributed
a. At 5% level significance tests the hypothesis that the two populations are
homogeneous (equal variances).
b. Assuming homogeneous populations, test the management claim using α=0.05
27
3.
A company claims that its medicine brand A provide faster relief from pain that
another company’s medicine brand B. a researcher tested both brands of medicines on
two groups of randomly selected patients. The results of the test are given in the
following table. The mean and the standard deviation of relief time are in minutes.
Brand
Sample size
Mean of relief time Standard deviation
of relief time
A
21
44
12.5
B
17
49
7.5
Assuming that relief time is normally distributed
a. Assuming equal variances test the company claim at 0.05 level of significance
b. Using α=0.05 test the hypothesis of homogeneous population (equal
variances)
28
Review:
1. In order to study the performance of CBA students in Stat. 120. The QMIS
department selected randomly 13 female and 12 male students and their final
scores are recorded giving the following summary statistics
Sample size
Mean
STD
Female
13
84.15
9.90
male
12
76.58
11.27
Assuming that the scores have homogeneous normal distributions test the
hypothesis that the female students scores on the average more than the male
students (α=0.05). (𝑯𝟏 : 𝝁𝟏 > 𝝁𝟐 , 𝒕𝒄 = 𝟏. 𝟕𝟖𝟖, 𝒕.𝟎𝟓,𝟐𝟑 = 𝟏. 𝟕𝟏𝟒, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
2. A sample of 18 fathers who were company executives showed that they spend an
average of 2.3 hours per week playing with their children, with a standard
deviation of 0.54 hours. Another sample of 24 fathers who were medical
professionals gave a mean of 4.6 hours per week with a standard deviation of 0.8
hours.
Assume that the times spent per week playing with their children by all fathers
who are executives and all fathers who are medical professional have normal
distributions with equal standard deviations.
a. Construct a 95% C.I for difference between the mean time spent per week
playing with their children by all fathers who are executives and all fathers
who are medical professionals. (-2.741, -1.858)
b. Using the above C.I, test whether the mean time spent by all fathers who
are executives is equal to that for all fathers who are medical
professionals. (𝑯𝟏 : 𝝁𝟏 ≠ 𝝁𝟐 , 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
3. A firm is studying the delivery times of two raw material suppliers. The firm is
basically satisfied with supplier A and is prepared to stay with that supplier if the
mean delivery time is the same as or less than that of supplier However, if the
firm finds that the mean delivery time of supplier B is less than that of supplier A,
it will begin making raw material from supplier B.
a. What are the null and alternative hypotheses for this situation?(𝑯𝟏 : 𝝁𝑨 > 𝝁𝑩 )
b. Assume the independent samples show the following delivery time for the
two suppliers:
Supplier A
Supplier B
𝑛1 =10
𝑛2 =20
𝑥̅1 =14 days
𝑥̅2 =12.5 days
𝑠1 =4
𝑠2 =2
With α=0.05 and using t-test with pooled variance what is your conclusion
for the hypothesis from part (a)? What do you recommend in terms of
supplier selection? (𝒕𝒄 = 𝟏. 𝟑𝟖, 𝒕.𝟎𝟓,𝟐𝟖 = 𝟏. 𝟕𝟎𝟏, 𝒅𝒐𝒏𝒐𝒕 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 ).
29
4. In a random sample of nine gasoline stations in City “A”, the prices per gallon of
unleaded gas have a standard deviation of $0.08 per gallon. In a random sample of
14 gasoline stations in city “B”, the prices per gallon have a standard deviation of
$0.03 per gallon. Use the 10% significance level to test the null hypothesis that
the price per gallon of gasoline is equally variable in two cities.
(𝑯𝟏 : 𝝈𝟐𝟏 ≠ 𝝈𝟐𝟐 , 𝑭𝒄 = 𝟕. 𝟏𝟏, 𝑭𝟎.𝟎𝟓,𝟖,𝟏𝟑 = 𝟐. 𝟕𝟕, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 ).
5. On the basis of data provided by a salary survey, the variance in annual salaries
for seniors in accounting firms is approximately 2.1 and the variance in annual
salaries for managers in accounting firms is approximately 11.1. The salary data
were provided in thousands of euros. Assuming the salary data were based on
sample of 25 seniors and 26 managers, test the hypothesis that the population
variances in the salaries are equal. At α=0.05, what is your conclusion?
(𝑯𝟏 : 𝝈𝟐𝟏 ≠ 𝝈𝟐𝟐 , 𝑭𝒄 = 𝟓. 𝟐𝟗, 𝑭𝟎.𝟎𝟐𝟓,𝟐𝟓,𝟐𝟒 = 𝟐. 𝟐𝟔, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 ).
30
III.
The difference between two population means (𝝁𝟏 − 𝝁𝟐 = 𝝁𝒅 ) for dependent
(paired) samples
 Point estimate of (𝜇1 − 𝜇2 = 𝜇𝑑 )
𝜇̂ 𝑑 = 𝑑̅
 Confidence interval C.I
𝑆
𝑑̅ ± 𝑡𝛼⁄2,𝑛−1 𝑑
𝑖𝑓 𝜎 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑎𝑛𝑑 𝑛 < 30
√𝑛
Where
 d: difference between the two variables

∑𝑑
𝑑̅ = 𝑛

𝑆𝑑2

𝑆𝑑 = √𝑆𝑑2
=
∑ 𝑑2−
(∑ 𝑑)2
𝑛
𝑛−1
or
𝑆𝑑2 =
∑ 𝑑2 −𝑛𝑑̅ 2
𝑛−1
or
𝑆𝑑2 =
∑(𝑑−𝑑̅)2
𝑛−1
 Hypothesis test
𝑇𝑐 =
𝑑̅ −0
𝑆𝑑
√𝑛
𝑖𝑓 𝜎 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑎𝑛𝑑 𝑛 < 30
𝑇𝑐 ~𝑡𝑛−1
Note:
 We will use Z instead of T in both C.I and hypothesis test if
o σ is known
o σ is unknown with n≥30
 there are three ways to do the test as mentioned in chapter 9 in this note
31
Exercises:
1. To test the difference between two body shop garages, 10 randomly chosen damaged cars
were sent to these two garages (A and B). The following are the estimated repair garages
of these garages.
A
236
137
379
255
279
321
369
333
137
390
B
310
187
392
232
321
318
389
288
167
432
d=A-B
𝒅𝟐
Assuming that the repair charges are normally distributed
a. Test the hypothesis that the repair charge at garage A is lower than that at garage
B, state your assumptions.
b. Construct a 95% C.I of the difference between the two means.
32
2. The manufacture of gasoline additive claims that the use of this additive increases
gasoline mileage. A random sample of 6 cars were driven for one week with the gasoline
additive and then for one week without the gasoline additive. The following table
provides the obtained information about the gasoline mileage.
Gasoline mileage
With
without
D=with-without
Mean
25.12
23.4
1.717
Standard deviation
5.87
5.42
1.427
a. Compute a 99% confidence interval for the mean difference gasoline mileage?
b. Is it possible to say that the manufacturer’s claim is true? Why? Use α=0.01
33
Review:
1. A company claims that its 12-week special exercise program significantly reduces weight. A random
sample of six persons was selected, and these persons were put on this exercise program for 12
weeks. The following table gives the weights (in pounds) of those six persons before and after the
program. Assume that the population of all paired differences is (approximately) normally distributed.
Before
After
180
183
195
187
177
161
221
204
208
197
199
189
a. Make a 95% confidence interval for the mean of the population paired differences, where
a paired difference is equal to the weight before joining this exercise program minus the
weight at the end of the 12-week program. (2.278, 17.382)
b. Using the 1% significance level, can you conclude that the mean weight loss for all
persons due to this special exercise program is greater than zero?(𝑯𝟏 : 𝝁𝒅 > 𝟎, 𝒕𝒄 =
𝟑. 𝟑𝟓, 𝒕.𝟎𝟏,𝟓 = 𝟑. 𝟑𝟔𝟓, 𝒅𝒐𝒏𝒐𝒕 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
2. A study used to test whether a training course is helpful for students to pass a mathematics course. To
evaluate the effectiveness of the training course, eight students test scores were compared before and
after taking the training course. The results are as follows
Scores
student
before
after
1
46
50
2
52
50
3
64
71
4
67
70
5
58
54
6
55
61
7
60
62
8
60
68
a. Compute a 90% confidence interval for the mean difference scores? (0.25, 5.75)
b. Is it possible to say that the training course is helpful? Why? (𝑯𝟏 : 𝝁𝒅 > 𝟎, 𝒕𝒄 =
𝟐. 𝟎𝟔, 𝒕.𝟎𝟓,𝟕 = 𝟏. 𝟖𝟗𝟓)
3. A company is considering installing new machines to assemble its products. The company is
considering two types of machines, but it will buy only one type. The company selected 11 assembly
workers and asked them to use these two types of machines to assemble products. The time in
minutes to assemble one unit of the product on each type of machine for each of these eleven workers
is recorded and given to company statistician who supplied the following information
Machine І
23
26
19
24
27
22
20
18
17
21
25
Machine П
21
24
23
25
24
28
24
21
17
26
23
Assuming normality, use a confidence interval for the difference between the average assembly
time for the two machines to test the hypothesis that the two machines are the same at α=0.05.
(𝑯𝟏 : 𝝁𝒅 ≠ 𝟎, (−𝟑. 𝟒𝟔𝟐, 𝟎. 𝟗𝟏𝟔), 𝒅𝒐𝒏𝒐𝒕 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
34
IV.
The difference between two population proportions (𝑷𝟏 − 𝑷𝟐 )
𝑃̂1 − 𝑃̂2
 Point estimate of 𝑃1 − 𝑃2
 Confidence interval C.I
(𝑃̂1 − 𝑃̂2 ) ± 𝑍𝛼⁄2 √
𝑃̂1 (1 − 𝑃̂1 ) 𝑃̂2 (1 − 𝑃̂2 )
+
𝑛1
𝑛2
 Hypothesis test
𝑍𝑐 =
(𝑃̂1 − 𝑃̂2 ) − 𝐷
1
1
√𝑃̅ (1 − 𝑃̅) ( + )
𝑛1 𝑛2
Where
Combined sample proportion
𝑋 +𝑋
𝑃̅ = 𝑛1 +𝑛2
1
2
or
̂
̂
𝑛 𝑃 +𝑛 𝑃
𝑃̅ = 1𝑛1+𝑛2 2
1
2
35
Exercises:
1. A company has two restaurants in two different areas in Kuwait. The company wants to
estimate the percentage of customers who thinks that the food and service at each of these
restaurants are excellent. A sample of 200 customers taken from restaurant in area A
showed that 118 think that the food and service are excellent in this restaurant. Another
sample of 250 customers taken from restaurant in area B showed that 160 think that the
food and service are excellent in this restaurant.
a. Find the point estimate of the difference between the two proportions.
b. Construct a 97% C.I of the difference between the two proportions.
c. Find the p-value to test the hypothesis that the proportion of customers who thinks
that the food and service in area A is lower than the corresponding proportion at
the restaurant in area B.
d. What is your conclusion if α=0.025?
36
Review:
1. In a random sample of 800 men aged between 25 to 35, 24% of them said they live with
one parent. In other sample of 850 women of the same age group, 18% said that they live
with one parent. Construct a 95% confidence interval for the difference between the two
population proportions. (0.021, 0.099)
2. A company that has many department stores wanted to find at two such stores the
percentage of sales for which at least one of the items was returned. A sample of 800
sales randomly selected from store A showed that for 240 of them at least one of the
items was returned. Another sample of 900 sales randomly selected from store B showed
that for 279 of them at least one of the items was returned.
a. Construct at 98% confidence interval for the difference between the proportions
of all sales at the two stores for which at least one of the items was returned.
(-0.0621, 0.0421)
b. Find the standard error and the margin error of C.I. (0.02236,0.05211)
c. Using the 1% significance level, can you conclude that at the two stores the
proportions of all sales for which at least one of the items was returned are
different?(𝑯𝟏 : 𝑷𝑨 ≠ 𝑷𝑩 , 𝒁𝒄 = −. 𝟒𝟓, 𝒅𝒐𝒏𝒐𝒕 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
d. Find the p-value for the test mentioned in part (c). (0.6528)
e. Find the standard error of the test. (0.02237)
37
Chapter 11
ANOVA
 Assumptions:
1. “k” random independent samples from
2. Normal population with
3. Equal variances (homogenous populations)
 Hypothesis test:
1. 𝐻0 : 𝜇1 = 𝜇2 = ⋯ = 𝜇𝑘
vs
𝐻1 : at least one population means is different
Where k: # of samples or groups or populations
2. T.S
𝐹𝑐 =
3.
𝑀𝑆𝐵 (𝑀𝑆𝐹)
𝑀𝑆𝑊 (𝑀𝑆𝐸)
Calculated from
ANOVA table
Determine the Critical value
𝐹𝛼,𝑘−1,𝑛−𝑘
4. Conclusion: Reject 𝐻0 if T.S (𝐹𝑐 ) lies in the rejection region R.R (under shaded area).
38
Source of
variation
Between
(Factor)
Degrees of freedom
(d.f)
k-1
Sum of squares
(SS)
SSB
(SSF)
Within
(Error)
n-k
SSW
(SSE)
Total
n-1
SST






SSW(SSE)
𝑇12
SSW(SSE)
n−k
Fc =
MSB(MSF)
MSW(MSE)
-
𝑇22
𝑇1 = 𝑛1 𝑥̅1 ,
𝑇𝑘2
𝑇2 = 𝑛2 𝑥̅ 2 , … , 𝑇𝑘 = 𝑛𝑘 𝑥̅𝑘
𝑇2
SSB = SST-SSW
SSB = MSB (k-1)
SSB = (
SSW = SST-SSB
SSW = (𝑛1 − 1)𝑆12 + (𝑛2 − 1)𝑆22 + ⋯ + (𝑛𝑘 − 1)𝑆𝑘2 = ∑(𝑛𝑖 − 1)𝑆𝑖2
SST = SSB+SSW
𝑀𝑆𝐸 ∗ = 𝑆𝑃2 =
𝑛1
+
𝑛2
+ ⋯+
SSB = ∑ 𝑛𝑖 (𝑥̅𝑖 − 𝑥̅ )2
SSW = MSW (n-k)
SST
MSW(MSE) =
k: number of samples/ groups/ populations.
𝑛 = 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 (𝑡𝑜𝑡𝑎𝑙 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒).
𝑇1 = ∑ 𝑥1 , 𝑇2 = ∑ 𝑥2 , … , 𝑇𝑘 = ∑ 𝑥𝑘
𝑜𝑟
𝑇 = 𝑇1 + 𝑇2 + ⋯ + 𝑇𝑘
𝑆12 , 𝑆22 , … , 𝑆𝑘2
∑ 𝑥 2 = ∑ 𝑥12 + ∑ 𝑥22 + ⋯ + ∑ 𝑥𝑘2
SSB(SSF)
Mean squares
(MS)
SSB(SSF)
MSB(MSF) =
k−1
𝑛𝑘
)−
𝑇2
𝑇2
𝑇2
1
2
𝑘
𝑛
SSW=∑ 𝑥 2 − (𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 )
SST =∑ 𝑥 2 −
𝑇2
𝑛
(𝑛1 − 1)𝑆12 + (𝑛2 − 1)𝑆22 + ⋯ + (𝑛𝑘 − 1)𝑆𝑘2
𝑛−𝑘
Note:
 ∑ 𝒙𝟐 ≠ (∑ 𝒙)𝟐
 𝑻𝒊 𝟐 ≠ ∑ 𝒙𝟐𝒊
 𝑻𝒊 𝟐 = (∑ 𝒙𝒊 )𝟐
39
Exercise
1. A consumer agency wanted to find out if the mean time it takes for each of three brands
of medicine to provide relief from a headache is the same. The three drugs were
administered to three randomly selected samples. The following table gives the time in
minutes taken by each patient to get relief from a headache, followed by a Minitab output
to such problem.
Drug
П
14
20
18
24
І
25
38
42
65
47
52
Level
Drug І
Drug П
Drug Ш
N
6
4
5
Mean
44.83
19.00
53.60
StDev
13.50
4.16
13.24
Ш
44
39
54
58
73
Individual 95% CIs for Mean
Based on Pooled StDev
-------+---------+---------+--------(------*------)
(-------*-------)
(-------*------)
-------+---------+---------+--------16
32
48
a. Complete the analysis of variance table
Source
Factor
Error
Total
df
SS
MS
F
b. Test the consumer agency claim at 5% level of significance
c. Suppose that the hypothesis of equal means has been rejected which of the drugs is
different and why?
40
2. A panel of trained testers judges the flavor quality of different vanilla frozen desserts:
frozen yogurts, ice milks, other frozen desserts measured on a scale from 0 to 100. The
sample sizes are respectively, 𝑛1 = 13, 𝑛2 = 8, 𝑎𝑛𝑑 𝑛3 = 6. Below is most of the
ANOVA output from the computer:
Source
Factor
(Between)
Error
(Within)
Total
df
?
SS
6364
MS
3182
24
3031
?
?
?
F
?
a. Complete the ANOVA table
b. Test whether there is a significant difference in the flavor quality of the three
different disserts
 State the null and alternative hypotheses

Find the value of the test statistic

Find the critical value(s). use 0.05 significance level

What is your conclusion about the flavor quality of the three different
disserts?

What are the assumptions required to make this test?
41
3. Sex hours were selected from each of 3 radio stations, and analysis of variance was
performed on the data. Part of the ANOVA table is shown below
Source
Between
Within
Total
df
SS
MS
1311.02
F
13.368
a. Complete ANOVA table
b. At 0.05 significance level, is there a difference in the stations means
42
Review:
1. A statistics professor has developed four methods (M1, M2, M3, M4) for teaching a
senior level class. He wishes to investigate if there is a difference in the four methods.
The professor assigns students to the four teaching methods. The final exam scores
for each group were recorded. The four sample sizes and sample means are in the
following table:
Method
Sample size
Sample mean
M1
7
79
M2
4
83.75
M3
6
70
M4
5
72.8
Also you are given that the error (within) sum of squares ”SSE” (SSW)=861.55
Carry out ANOVA test using a 1% level of significance
 State the null and alternative hypotheses
 Find the value of test statistic (3.973)
 Find the critical value (5.09)
 What is your conclusion about the four different methods of teaching? (don’t
reject 𝑯𝟎 ).
Source
Between
Within
Total
df
3
18
21
SS
570.45
861.55
1432
MS
190.15
47.864
F
3.973
2. Samples were selected from three populations, the data obtained is given below
Sample 1
Sample 2
Sample 3
91
77
88
98
87
75
107
84
73
102
95
84
85
75
82
a. State the assumption needed to use the analysis of variance to test the equality
of the three population means
b. Test the hypothesis of no difference between the three population means at
0.05 level of significance.(𝑭𝒄 = 𝟏𝟏. 𝟖𝟖𝟓, 𝑭.𝟎𝟓,𝟐,𝟏𝟐 = 𝟒. 𝟒𝟔, 𝒓𝒆𝒋𝒆𝒄𝒕 𝑯𝟎 )
Source
Between
Within
Total
df
2
12
14
SS
968.7
489
1457.7
MS
484.35
40.75
F
11.885
43
Chapter 12
Independence
1. 𝐻0 : two variables are independent (𝐧𝐨𝐭 𝐫𝐞𝐥𝐚𝐭𝐞𝐝)
vs
𝐻1 : two variables are dependent (𝐫𝐞𝐥𝐚𝐭𝐞𝐝)
2. Calculate the test statistic (T.S)
𝑥𝑐2 = ∑
(𝑂 − 𝐸)2
𝐸
Where
O: Observed value
E: Expected value
E=
row total ∗ column total
total
3. Determine the critical value
2
𝑥𝛼,(𝑟−1)(𝑐−1)
4. Conclusion:
Reject 𝐻0 if T.S (𝑥𝑐2 ) lies in the rejection region (under shaded area).
44
Exercises:
1. Let's try an example. 500 elementary school boys and girls are asked which is their
favorite color: blue, green, or pink? Results are shown below:
Boys
Girls
Total
Blue
100
20
120
Green
150
30
180
Pink
20
180
200
Total
300
200
500
would you conclude that there is a relationship between gender and favorite color?
a. The two hypothesis
𝐻0 :
𝑣𝑠
𝐻1 :
b. The test statistic
c. The critical value(s). Use 0.05 significance level.
d. The conclusion
45
2. One hundred auto drivers who were stopped by police for some violation were
also checked to see if they were wearing seat belt. The following table records
the results of this survey
Wearing seat belt
Not wearing seat belt
Total
Men
34
21
55
Women
32
13
45
total
66
34
100
For a chi square test of independence for this contingency table:
a. What is the number of degrees of freedom?
b. What is the total of the second row?
c. How many drivers are in the sample ?
d. What are the observed frequencies for the first row?
e. What are the expected frequencies for the second row?
f. What are the observed frequencies for the second column?
g. What are the expected frequencies for the second column?
46
CHAPTER 13
Simple Linear Regression
I.
The population regression model
𝑦 = 𝛽0 + 𝛽1 𝑥 + 𝜀
Where
o
o
o
o
o
𝑦: is the dependent variable
𝑥: is the independent variable
𝛽0: is y-intersection or constant term
𝛽1: is the slope
𝜀: is a random error term
 Estimate the population regression model by the sample linear regression model
𝑦̂ = 𝑏0 + 𝑏1 𝑥
This equation is called the least squares regression line or the prediction equation.
 Sum of squares
𝑆𝑆𝑥𝑦 = ∑ 𝑥𝑦 −
∑𝑥∑𝑦
𝒐𝒓 𝑆𝑆𝑥𝑦 = ∑ 𝑥𝑦 − 𝑛𝑥̅ 𝑦̅ 𝒐𝒓 𝑆𝑆𝑥𝑦 = ∑(𝑥 − 𝑥̅ )(𝑦 − 𝑦̅)
𝑛
𝑆𝑆𝑥𝑥
(∑ 𝑥)2
= ∑𝑥 −
𝒐𝒓 𝑆𝑆𝑥𝑥 = ∑ 𝑥 2 − 𝑛𝑥̅ 2 𝒐𝒓 𝑆𝑆𝑥𝑥 = ∑(𝑥 − 𝑥̅ )2
𝑛
𝑆𝑆𝑦𝑦
(∑ 𝑦)2
= ∑𝑦 −
𝒐𝒓 𝑆𝑆𝑦𝑦 = ∑ 𝑦 2 − 𝑛𝑦̅ 2 𝒐𝒓 𝑆𝑆𝑦𝑦 = ∑(𝑦 − 𝑦̅)2
𝑛
2
2
 Estimated value of 𝛽0 and 𝛽1
𝑆𝑆𝑥𝑦
𝑆𝑆𝑥𝑥
̂
𝐵0 = 𝑏0 = 𝑦̅ − 𝑏1 𝑥̅
𝐵̂1 = 𝑏1 =
 Prediction value of y
𝑦̂ = 𝑏0 + 𝑏1 𝑥
given
 Residual (Error)
𝑒 = 𝑦 − 𝑦̂
given
47
II.
How to Evaluate the estimated model
1. Coefficient of Determination
𝑟2 =
𝑏1 𝑆𝑆𝑥𝑦 𝑆𝑆𝑅
=
𝑆𝑆𝑦𝑦
𝑆𝑆𝑇
0 ≤ 𝑟2 ≤ 1
It explains the variation in “y” by the independent variable “x”
𝑟 2 increased
Note: SSR increased
Good model
2. Coefficient of Correlation
𝑟 = √𝑟 2
or
𝑟=
(with the same sign of 𝑏1 )
𝑆𝑆𝑥𝑦
√𝑆𝑆𝑥𝑥 𝑆𝑆𝑦𝑦
= 𝑏1 √
𝑆𝑆𝑥𝑥
𝑆𝑆𝑦𝑦
−1 ≤ 𝑟 ≤ 1
It measures the strength of the linear relationship between two variables
Perfect
Perfect
III.
Estimation of the variance and standard deviation of random errors


The estimated variance of errors 𝜎̂𝜖2 = 𝑆𝑒2 = 𝑀𝑆𝐸
The estimated St. Deviation of errors 𝜎̂𝜖 = 𝑆𝑒 = √𝑀𝑆𝐸
𝑴𝑺𝑬 =
𝑺𝑺𝒚𝒚 − 𝒃𝟏 𝑺𝑺𝒙𝒚
𝑺𝑺𝑬
=
𝒏−𝒌
𝒏−𝒌
(k: number of parameters)
48
IV.
Inferences about 𝜷𝟏
𝒃𝟏 ~𝑵(𝝁𝒃𝟏 = 𝑩𝟏 , 𝑺𝒃𝟏 =

𝑺𝒆
√𝑺𝑺𝒙𝒙

Population simple linear regression equation
𝑦 = 𝛽0 + 𝛽1 𝑥 + 𝜀
Point estimate of 𝛽1: 𝛽̂1 = 𝑏1

Confidence interval of 𝛽1
o If 𝑛 < 30

Hypothesis test about 𝐵1
𝑏1 ± 𝑡𝛼⁄2,𝑛−𝑘 𝑆𝑏1
)
(k: number of parameters)
(Test of 𝑏1 or Test if there is a good relationship between
x and y).
1.
𝑯𝟎 : 𝜷 𝟏 = 𝟎
Means there is no relationship between X
and Y (not significant linear relationship)
(X will dropped from model)
𝑯𝟏 : 𝜷 𝟏 ≠ 𝟎
Means there is a relationship between X
and Y (significant linear relationship)
𝑯𝟏 : 𝜷 𝟏 > 𝟎
Positive linear relationship
𝑯𝟏 : 𝜷 𝟏 < 𝟎
negative linear relationship
2.Calculate Test statistic
𝑏
𝑇𝑐 = 𝑆 1
o If 𝑛 < 30
𝑏1
3.Determine the rejection region
−𝑡𝛼⁄2,𝑛−𝑘
𝑡𝛼⁄2,𝑛−𝑘
−𝑡𝛼,𝑛−𝑘
𝑡𝛼,𝑛−𝑘
4.Conclusion: Reject 𝐻0 if T.S lies in rejection region (R.R)
49
V.
Testing the overall model
1.
𝐻0 : 𝛽1 = 0
The model is not significant/ not useful/not good/not
adequate /data not fit model.
𝐻1 : 𝛽1 ≠ 0
The model is significant/ useful/ good/adequate/data
fit model
2. Test statistic
𝑀𝑆𝑅
𝐹𝑐 = 𝑀𝑆𝐸
(Calculated from ANOVA table)
3. Critical value
𝐹𝛼,𝑘−1,𝑛−𝑘
4. Conclusion
Reject 𝐻0 if T.S (𝐹𝑐 ) lies in the rejection region R.R (under shaded area).
ANOVA table
Source of
variation
Regression
Residual
Error
Total
d.f
Sum of squares SS
Mean squares MS
k-1
𝑆𝑆𝑅 = 𝑏1 𝑆𝑆𝑥𝑦
n-k
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝑅
𝑆𝑆𝑅
𝑘−1
𝑆𝑆𝐸
𝑀𝑆𝐸 =
𝑛−𝑘
n-1
𝑆𝑆𝑇 = 𝑆𝑆𝑦𝑦
𝑀𝑆𝑅 =
T.S
𝐹𝑐 =
𝑀𝑆𝑅
𝑀𝑆𝐸
𝒌 = # 𝒐𝒇 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓𝒔
50
CHAPTER 14
Multiple Regressions
I.
Least squares regression line equation
𝑦̂ = 𝑏0 + 𝑏1 𝑥1 + 𝑏2 𝑥2 + 𝑏3 𝑥3 + ⋯
𝑦̂: Dependent variable (prediction of y)
𝑥𝑖 : Independent variable
II.
Hypothesis test
1. 𝐻0 : 𝛽1 = 𝛽2 = 𝛽3 = ⋯ = 0 (the model is not significant)
Vs
𝐻1 : at least one β is not equal to zero (the model is significant)
2. Test statistic
𝑀𝑆𝑅
𝐹𝑐 = 𝑀𝑆𝐸
(Calculated from ANOVA table)
3. Critical value
𝐹𝛼,𝑘−1,𝑛−𝑘
4. Conclusion
Reject 𝐻0 if T.S (𝐹𝑐 ) lies in the rejection region R.R (under shaded area).
ANOVA table
Source of
d.f
variation
Regression
k-1
Residual
Error
Total
Sum of squares SS
Mean squares MS
𝑆𝑆𝑅 = 𝑏1 𝑆𝑆𝑥𝑦
𝑆𝑆𝑅
𝑘−1
𝑆𝑆𝐸
𝑀𝑆𝐸 =
𝑛−𝑘
n-k
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝑅
n-1
𝑆𝑆𝑇 = 𝑆𝑆𝑦𝑦
𝑀𝑆𝑅 =
T.S
𝐹𝑐 =
𝑀𝑆𝑅
𝑀𝑆𝐸
51
Exercises:
1. The following table gives information on the temperature in a city and the volume of ice
cream (in pounds) sold at an ice cream parlor for a random sample of eight days during the
summer of 1999.
Temperature
93
86
77
89
98
102
87
79
Ice cream sold
208
175
123
198
232
277
158
117
∑𝑥 = 711 , ∑𝑦 = 1488, ∑𝑥 2 = 63713, ∑𝑦 2 = 297428, ∑𝑥𝑦 = 135466, 𝑥̅ = 88.88, 𝑦̅ = 186
a. Find sum of squares (SS)
b. Find the least squares regression line (𝑦̂ = 𝑎 + 𝑏𝑥)
c. Give a brief interpretation of the values of a and b
-
a(𝑏0 ): the initial value of y when x=0 (the initial value of the volume of ice-cream sold (y) is
equal to a= -361.5008 when the temperature (x)is equal to zero)
-
b(𝑏1 ): if we increase x by 1 unit then y will change (increase or decrease) by the value of b. (if
we increase the temperature (x) by 1 degree then the volume of ice-cream sold (y) will increase
by b=6.16.
d. Compute 𝑟 and 𝑟 2 , explain what they mean
e. Predict the amount of ice cream sold on a day with a temperature of 95°
f. Compute the standard deviation of errors
g. Construct a 99% confidence interval for 𝛽1
h. Testing at 1% significance level, can you conclude that 𝛽1 is different from
zero?
52
2. Regression analysis was applied between sales data (in $1000) and advertising data
(in $100) and the following information was obtained
𝑦̂ = 12 + 1.8𝑥
𝑛 = 17, 𝑆𝑆𝑅 = 225, 𝑆𝑆𝐸 = 75, 𝑆𝑏1 = 0.2683
a. Based on the above estimated regression equation, if advertising is $3000,
then the predicted value of sales (in dollars) is
b. The F statistic computed from the above data is
c. The critical F value at α=0.05 is
d. Is the estimated regression model significant?
 The two hypotheses
 Your conclusion
e. The t statistic for testing the significance of the slope is
f. Is the linear relationship between X and Y significant? Use the t-test to answer
this equation.
 The two hypotheses
 The critical value(s)
 Your conclusion
g. Calculate the 95% confidence interval of the slope of the regression line for
all statistics students
h. Develop an analysis of variance table
Source of
d.f
SS
variation
MS
53
3. The owner of a bowling establishment is interested in the relationship between the
price she charges for a game of bowling and the number of games bowled per day.
She collected data on the number of games bowled per day at 15 different prices. Fill
in the missing entries in the following MINITAB output that was obtained for these
data. In this output, X represents the price of a game and Y is the number of games
bowled per day.
The regression equation is
y = ____ ____ x
Predictor
Constant
x
Coef
691.02
-141.30
S = _____
STDEV
21.70
____
R-Sq = _____
T
____
-16.83
P
0.000
0.000
R-Sq(adj) = 95.3%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
___
___
___
SS
148484.0
_______
_______
MS
_______
_______
F
283.13
P
0.000
4. A researcher wanted to examine the relation between a dependent variable y and an
independent variable x, He selected randomly 10 observations giving the following partial
MINITAB output:
A.
Predictor
Constant
X1
S = 15.2158
Coef
247.97
-8.172
SE Coef
15.01
1.077
R-Sq = 87.8%
R-Sq(adj) = 86.3%
1. Write down the estimated regression equation:
2. Test the hypothesis of no regression (𝑏1 =0) using α=0.05
3. Find the correlation coefficient between y and x1 and comment on its value
54
B. To researcher decided to add another independent variable x2 to the model in (A)
he obtained the following results
The regression equation is
y = 180 - 6.17 x1 + 0.848 x2
Predictor
Constant
X1
X2
S = 10.2957
Coef
180.07
______
0.8484
SE Coef
23.31
______
0.2622
R-Sq = _______
T
_____
-6.46
_____
R-Sq(adj) = 93.7%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
2
_
9
SS
______
742.0
15182.9
MS
_______
_______
F
_____
1. Complete the missing values in the previous MINITAB output
2. Test at α=0.05 of no regression model (i.e. the overall model does not fit the
data).
3. Find a 95% confidence interval for the regression coefficient of x2
4. Test the hypothesis that the true value of the regression coefficient of x1 equal
to 0
5. Which of the two models you prefer, the one estimated in (A) above or the
one estimated in (B) above? And why?
6. Test the hypothesis that the true value of the regression coefficient of 𝑥2 is positive
(α=.05).
55
Inference about 𝜷𝒊
The regression equation is
y = 180 - 6.17 x1 + 0.848 x2
𝑏0
𝑏1
Predictor
Constant
𝑏2
Coef
SE Coef
𝑏0 180.07
X1
𝑏1 -6.17
X2
𝑏2
T
𝑆𝑏0 23.31
7.725
𝑆𝑏2 0.2622
S = 10.2957
R-Sq = 0.9511
Estimated standard
deviation of error
Coefficient of
determination
√𝑀𝑆𝐸
𝑆𝑆𝑅
𝑆𝑆𝑇
0.00
𝑏𝑖
-6.46
𝑆𝑏𝑖
𝑆𝑏1 0.955
0.8484
p-value
0.00
3.236
0.00
Testing hypothesis for
two tailed test we
compare p-value with
α. The significant of
𝛽𝑖 or the linear
relationship between
𝑥𝑖 𝑎𝑛𝑑 𝑦𝑖
R-Sq(adj) = 93.7%
Over all model (conduct the f test of model usefulness/test the whole model/
𝑯𝟎 : 𝜷𝟏 = 𝜷𝟐 = 𝟎 𝒗𝒔 𝑯𝟏 : 𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝒐𝒏𝒆 𝜷 ≠ 𝟎)
Analysis of Variance
Source
DF
SS
Regression
2
14440.9
Residual Error
7
742.0
Total
9
15182.9
MS
F
7220.45
68.117
106
Error mean square
𝑀𝑆𝐸 =
Or
𝑆𝑆𝐸
𝑛−𝑘
𝑀𝑆𝐸 = 𝑆 2
56
5. The wner of ShowTime Movie theaters would like to estimate weekly gross revenue as a
function of advertising expenditure. Historical data for a sample of 8 weeks follow
Weekly gross Revenue (Y)
Television advertising (X1)
Newspaper advertising (X2)
($1000s)
($1000s)
($1000s)
96
5.0
1.5
90
2.0
2.0
95
4.0
1.5
92
2.5
2.5
95
3.0
3.3
94
3.5
2.3
94
2.5
4.2
94
3.0
2.5
A portion of the MINITAB computer follows
The regression equation is
y = 83.2 + 2.29 x1 + 1.30 x2
Predictor
Constant
X1
X2
S = 0.642587
Coef
SE Coef
1.574
0.3041
0.3207
T
R-Sq = 91.9%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
SS
23.435
MS
F
5
a. What is the estimate of the weekly gross revenue for a week when $3500 is spent
on television advertising and $1800 is spent on newspaper advertising?
57
b. Find and interpret 𝑅 2
c. When television advertising was the only independent variable, 𝑅 2 =0.653
(65.3%) do you prefer the multiple regression results? Why?
d. Use α=0.05 to test the hypotheses 𝐻0 : 𝛽1 = 𝛽2 = 0, 𝐻1 : 𝛽1 𝑎𝑛𝑑/𝑜𝑟 𝛽2 is not equal
to zero. Did the estimated regression equation provide a good fit to the data?
Explain
e. Find the mean square error. Find the standard error of the estimate (𝜎̂)
f. Use α=0.05 to test the significance of each independent variable. Should X1 or X2
be dropped from the model
58
TABLES
59
60
61
62
63
64
65
66
67
68
69
70