qt ii (hy i) & (hy ii)
DESCRIPTION
Qt II (Hy i) & (Hy II)TRANSCRIPT
Hypothesis Testing
Steps of Hypothesis Testing
Setp-1. Establish a null and alternative hypothesis.
Step 2. Determining the appropriate statistical test.
Step 3. Set the value of alpha, the type I error rate.
Step 4. Establish the decision rule.
Step 5. Gather sample data.
Step 6. Analyze the data.
Step 7. Reach a statistical conclusion.
Step 8. Make a business decision.
Null and the Alternative Hypothesis.
The null hypothesis asserts that there is no true difference in the sample statistics and the population parameter under consideration and the difference found is accidental arising out of fluctuation of the sampling.
A hypothesis which states that there is no difference between assumed and actual value of the parameter is called the null hypothesis and the hypothesis that is different from the null hypothesis is the alternative hypothesis. Null hypothesis is denoted by and alternative hypothesis by .
Example: An auditor wishes to test the assumption that the mean value of all accounts receivable in a given firm is $260.00 by taking a sample of n=36 and computing the sample mean. He wishes to reject the assumed value of $260.00 only if it is clearly contradicted by the sample mean, and thus the hypothesized value should be given the benefit of doubt in the testing procedure. The null and the alternative hypothesis of the procedure are
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00.260$ ,00.260$ 10 HandH
Type I and Type II error
Statistical decisions generally entail some risk of error. We define two types of error
Error Meaning Risk Symbol
Type I Rejecting the null hypothesis when it is true
Type II Accepting the null hypothesis when it is false
Reject Type I Error Correct
Accept Correct Type II Error
Decision is either to
Null hypothesis is either
True False
Null and Alternative Hypotheses : Example
A soft drink company is filling 12 oz. Cans with cola.
The company hopes that the cans are averaging 12 ounces.
OZh 12:0 OZha 12:
Rejection and non rejection region
Statistical outcomes that result in the rejection of the null hypothesis lie in what is termed the rejection region. Statistical outcomes that fail to result in the rejection of the null hypothesis lie in what is termed as non rejection region.
=40 ounces
Non Rejection Region
Rejection Region
Critical Value
Rejection Region
Critical Value
One tail and two tail test
40:
40:
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40:
40:
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=40 ounces
Rejection Region
Non Rejection Region
Critical Value
=40 ounces
Rejection Region
Non Rejection Region
Critical Value
=12 OZ
Rejection Region
Non Rejection Region
Critical Value
Rejection Region
12:
12:
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Two tail test.
Testing of Hypothesis about a population mean using the z-Statistic
Formulas for testing hypothesis
1. Formula below can be used to test hypothesis about a single population mean if the sample size is large (n ) for any population and for samples (n<30) if x is known to be normally distributed
30
n
xz
2. When population standard deviation is unknown
ns
xz
3. Testing the Mean with a Finite Population
1
NnN
n
xz
P-Value Method to Test Hypothesis
-- No preset value of is given in the p-value method
-- The probability of getting a test statistic at least as extreme as the observed test statistic (computed from the data) is computed under the assumption that the null hypothesis is true.
-- p value is the smallest value of for which the null hypothesis can be rejected.
Procedure for two tailed test
Split alpha to determine the critical value of the test statistic.
P, the probability of getting a test statistic at least as extreme as observed value is computed.
If Do not reject H0
Example: If the p-value of a test is .038, the null hypothesis cannot be rejected at
=0.01 because .038 is the smallest value of for which the null hypothesis can be rejected. However, the null hypothesis can be rejected for =0.05.
P2
Exercise 9.5 (Page 306)
According to the U.S. Bureau of Labor Statistics, the average weekly earnings of a production worker in 1997 were $424.20. Suppose a labor researcher wants to test to determine whether this figure is still accurate today. The researcher randomly selects 54 production workers from across the United States and obtains a representative earning statement for one week from each. The resulting sample average is $432.69, with a standard deviation of $33.90. Use this data and hypothesis testing techniques along with a 5% level of significance to determine whether the mean weekly earnings of a production worker have changed. Also solve the problem by using the p-value method.
Testing Hypothesis about a Population Mean Using the t-statistic
The formula for testing such hypothesis
1
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t
Example 9.17
Suppose that in the past years the average price per square foot for warehouses in the United States has been $32.28. A national real state investor wants to determine whether the figure has changed now. The investor hires a researcher who randomly samples 19 warehouses that are for sale across the United States and find that the mean price per square foot is $31.67, with a standard deviation of $1.29. If the researcher uses a 5% level of significance, what statistical conclusion
can be reached? What are the hypothesis.
Ex. 9.16 Suppose a study report that the average price for a gallon of self-serve regular unleaded gasoline is $1.16.You believe that the figure is higher in your area of the country. You decide to test this claim for your part of the United States by randomly calling gasoline stations. Your random survey of 25 stations produces the following prices.
$1.27 $1.29 $1.16 $1.20 $1.37
1.20 1.23 1.19 1.20 1.24
1.16 1.07 1.27 1.09 1.35
1.15 1.23 1.14 1.05 1.35
1.21 1.14 1.14 1.07 1.10
Assume gasoline prices for a region are normally distributed. Do the data you obtained provide enough evidence to reject the claim? Use a 1 % level of significance.
Testing Hypothesis about a Proportion
p-1q ,proportion populationp
proportion ˆ
.ˆ
samplepwherenqppp
z
Example 9.22. The Independent Insurance Agents of America conducted a survey of insurance consumers and discovered that 48% of them always reread their insurance policies, 29% some time do, 16% rarely do, and 7% never do. Suppose a large insurance company invests considerable time and money in rewriting policies so that they will be more attractive and easy to read and understand. After using the new policies for a year, company managers wants to determine whether rewriting the policies significantly change the proportion of the policy holders who always reread their insurance policy. They contact 380 of the company’s insurance consumers who purchased a policy in the past year and ask them whether they always reread their insurance policies. One hundred and sixty four respond that they do. Use a 1% level of significance to test the hypothesis.
Solving for Type II Errors (Determine the Probability of Committing a Type II Error)
-- A type II error can be committed only when the researcher fails to reject the null hypothesis and the null hypothesis is false
-- A type II error , ,varies with the possible values of the alternative parameters
Procedure for computing the type-II error
Calculate the critical value for the sample mean
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10.0)645.1(12
nZX cC
0,979.11 HrejectxIf
0,979.11 HacceptxIf
Figure in the next slide gives distribution of values when the null hypothesis is true (top) and alternative mean ounces is true (bottom)
How often will the business researcher fail to reject the top distribution as true when, in reality, the bottom distribution is true
99.11
A soft drink company is filling 12 oz. Cans with cola
The company hopes that the cans are averaging 12 ounces.
OZH 12:0
OZH a 12:
645.116.1
60
10.00.12985.11
05.0
ZZ
Suppose a sample of 60 cans of beverage yields a sample mean of 11.985 ounces with a standard deviation of 0.10 ounces. and a one tailed test.
05.0
Decision : Do not reject null hypothesis
Alternative : Either correct decision or type –II error
If mu actually equals 11.99 ounces, what is the probability of failing to reject mu=12 ounces when 11.979 ounces is the critical value?
85.0
60
10.99.11979.111
1
n
sX
Z c
The value of Z yields an area of .3023. The probability of committing type-II error is the area to the right of in the lower distribution, or 0.3023+0.5=0.8023.
Fail to reject the null hypothesis but is correct.
979.11cx
Hence there is an 80.23 % chance of committing a type-II error if the alternative mean is 11.99 ounces.
12 99.11
11.999 .94 .06
11.995 .89 .11
11.990 .80 .20
11.980 .53 .47
11.970 .24 .76
11.960 .07 .93
11.950 .01 .99
Values and Power Values for the Soft-Drink Example
Power : Power is the probability of rejecting the null hypothesis when it is false and representing the correct decision of selecting the alternative hypothesis when it is true.Power is equal to
Value of for various values of alternative means
1
Power
Operating characteristic curve : Plotting the against the various values of the alternative hypothesis.
Power Curve : Plotting the power values against the various values of the alternative hypothesis.
Exercise 9.37 (Page 330)
The New York Stock Exchange recently reported that the average age of a female shareholder is 44 years. A broker in Chicago wants to know whether this figure is accurate for the female shareholders in Chicago. The broker secures a master list of share holders in Chicago and takes a random sample of 58 women. Suppose the average age of the shareholders in the sample is 45.1 years, with a standard deviation of 8.7 years. Test to determine whether the brokers sample data differ significantly enough from the 44 year figure released by the New York Stock Exchange to declare that the Chicago female shareholders is different in age from female shareholders in general. Use alpha=0.5. If no significant difference is noted, what is the brokers probability of committing a type II error if the average age of a female shareholder is actually 45 years? 46 years? 47 years? 48 years? Construct an OC curve for the data.Construct a power curve for the data.
A small business has 37 employees. Because of the uncertain demand for its product, the company usually pays overtime on any given week. The company assumed that about 50 total hours of overtime per week is required and that the variance on this figure is about 25. Company officials want to know whether the variance of overtime hours has changed. Given here is a sample of 16 weeks of overtime data in hours per week. Assume hours of overtime are normally distributed. Use these data to test the null hypothesis that the variance of overtime data is 25. Let alpha=0.10.
57 56 52 44
58 53 44 44
48 51 56 48
63 53 51 50
9.30 A manufacturing company produces bearings. One line of bearings is specified to be 1.64 centimeters (cm) in diameter. A major customer requires that the variance of the bearings be more than .001 cm2. The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise instrument, resulting in the following values. Assume bearing diameters are normally distributed. Use the data and to test the data to determine whether the population of these bearings is to be rejected because of too high a variance.
1.69 1.62 1.63 1.70
1.66 1.63 1.65 1.71
1.64 1.69 1.57 1.64
1.59 1.66 1.63 1.65
01.
Statistical Inferences about Two Populations
In some research design, the sampling plan calls for selecting two independent samples.
The object might be to determine whether the two samples come from the same population or if they come from the different populations, to determine the amount of difference in the populations.
Examples : Whether the effective ness of two brands of toothpaste differs or whether two brands of tires wire differently.
A business analyst might want to compare the expenditures on shoes made in 1992 with those from 2002 in an effort to determine whether any change occurred over time.
Hypothesis Testing and Confidence Intervals About the Difference in Two Means Using the Z – Statistic.
The Central Limit Theorem states that the difference in two sample means, is normally distributed for large sample sizes ( both and ) regardless of the shape of the populations.
It can be shown that
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2
22
1
12
21 nnxx
Z formula for the difference in two sample means for and
(independent Samples )
the mean of pop1
the mean of pop1
the mean of pop1
the mean of pop1
If the populations are normally distributed on the measurement being studied and if the population variances are known, above formula can be used for small sample sizes
Formula when population variances are unknown and samples are large
1n 302 n
2
22
1
12
2121
nn
xxz
1
2
1n1n
2
22
1
12
2121
nS
nS
xxz
Confidence Intervals
Confidence intervals for the difference in two populations means
When population standard deviations are unknown and large samples then the formula is
2
22
1
12
21212
22
1
12
21 nnzxx
nnzxx
2
22
1
12
21212
22
1
12
21 n
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szxx
n
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szxx
Exercise 10.6
The Bureau of Labor Statistics shows that the average insurance cost to a company per employee per hour is $ 1.84 for managers and $ 1.99 for professional specialty workers. Suppose these figures were obtained from 35 managers and 41 professional specialty workers and their respective standard deviations are $ 0.38 and $ 0.51. Calculate 98% confidence interval to estimate the difference in the mean hourly company expenditures for insurance for these two groups. What is the value of the point estimate? Test to determine whether there is a significant difference in the hourly rates employers pay for insurance between managers and professional specialty workers. Use 2% level of significance.
Exercise 10.7 :
A company’s auditor believes the per diem cost in Nashville, Tennessee, rose significantly between 1992 and 1999. To test this belief, the auditor samples 51 business trips from company’s records for 1992, the sample average was $ 190 per day, with a samples standard deviation of $ 18.50. The auditor selects a second random sample of 47 business trips from company’s record for 1999; the samples average was $ 198 per day, with a standard deviation of $ 15.60.If he uses a risk of committing a Type-I error of 0.01, does the auditor find that the per diem average expense in Nashville has gone up significantly?
Hypothesis Testing and Confidence Intervals About the Difference in Two Means :
Small Independent Samples and Population Variances Unknown
• Each of the two populations is normally distributed.
• The two samples are independent.
• At least one of the samples is small,
• The Values of the populations variances are unknown.
• The variances of the two populations are equal.
• The t formula to test the difference in means assuming is
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12
22
12
2
112
11
21
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122
112
2121
nndf
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When the population variances are not assumed to be equal
Exercise 10.17 (page 360). Based on an indication that mean daily car rental rates may be higher for Boston than in Dallas, a survey of eight car rental companies in Boston is taken and the sample means car rental rate $ 47, with a standard deviation of $ 3. Further, suppose a survey of nine car rental companies in Dallas results in a sample mean of $ 44 and a standard deviation of $3. Use alpha =0.01 to test to determine whether the average daily car rental rates in Boston are significantly higher than those in Dallas. Assume car rental rates are normally distributed and the population variances are equal.
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1
1
12
2
22
1
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11
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ns
n
ns
df
ns
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Statistical Inferences for Two Related or Dependent Samples
Formula to test hypothesis for dependent populations.
Where
N= number of pairs
d=sample difference in pairs
D=mean population difference
Sd=s.d. of samples difference
D=mean sample difference
Formula for and Sd
,
1
ndfn
sDd
td
d
n
dd
11
2
22
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dd
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ddsd
Exercise 10.27 (Page 368)
Eleven employees were put under the care of the company nurse because of high cholesterol readings. The nurse lectured them on the dangers of this condition and put them on the new diet. Shown are the cholesterol readings of the 11 employees both before the new diet and one month after use of the diet began. Construct a 98% confidence interval to estimate the population mean difference of cholesterol readings for people who are involved in this program. Assume differences in cholesterol readings are normally distributed in the population.
Employee Before After
1 255 197
2 230 225
3 290 215
4 242 215
5 300 240
6 250 235
7 215 190
Employee Before After
8 230 240
9 225 200
10 219 203
11 236 223
Use an alpha of 0.02 to test to determine whether there is a significant difference in the cholesterol readings.
Statistical Inferences about two population proportions, P1-P2
Applications
1. Comparing market share of a product for two different markets.
2. Comparing the proportions of defective products from one period to another.
3. Studying the difference in the proportion of female customers in two different regions.
For large samples
1
2
3
4
The difference in sample proportions is normally distributed with
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Z Formula for the Difference in Two Population Proportions
= Proportion from samples 1
= Proportion from samples 2
= size of sample 1
= size of sample 2
= Proportion from population 1
= Proportion from population 2
Q1 =1-P1
Q2=1-P2
2
22
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11
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ppppZ
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Z Formula to Test the Difference in Population Proportions
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Confidence Interval to Estimate P1-P2
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Exercise 10.35. Companies that recently developed new products were asked to rate which activities are most difficult to accomplish with new products. Options included such activities as assessing market potential, market testing, finalizing the design, developing a business plan, and the like. A researcher wants to conduct a similar study to compare the result between the two industries; the computer hardware industry and the banking industry. He takes a random sample of 56 computer firms and 89 banks. The researcher asks whether market testing is most difficult activity to accomplish in developing a new product. Some 48% of the sampled computer company and 56% of the sampled banks respond that it is most difficult activity. Use a level of significance of 0.20 to test whether there is a significant difference in the responses to the question from these two industries.
Exercise 10.36.A large production facility uses two machines to produce a key part for its main product.Inspectors have expressed concern about the quality of the finished product. Quality control investigation has revealed that the key part made by the two machines is defective at times. The inspectors randomly sampled 35 units of the key part from each machine. Of those products by machine A, five were defective. Seven of the 35 sampled parts from machine B were defective. The production manager is interested in estimating the difference in proportions of the populations of parts that are defective between machine A and machine B.From the sample information, compute a 98% confidence interval for this difference.
10.41 Suppose the data shown here are the result of a survey to investigate gasoline prices. Ten service stations were selected randomly in each of the two cities and the figures represent the prices of a gallon unleaded regular gasoline on a given day. Use the F test to determine wheher there is a significant difference in the variances of the prices of unleaded regular gasoline between two cities. Let =1.0. Assume gasoline prices are normally distributed.
City 1 City2
1.18 1.07 1.13 1.08 1.05 1.19
1.15 1.14 1.13 1.17 1.21 1.12
1.14 1.13 1.03 1.14 1.14 1.13
1.09 1.11
Exercise 10.42. How long are resale houses on the market? One survey by the Houston Association of realtors reported that in Houston, resale houses are on the market an average 112 days. Of course, the length of time varies by market. Suppose random samples of 13 houses in Houston and 11 houses in Chicago that are for resale are traced. The data shown here represent the number of days each house was on the market before being sold. Use the given data and a 1 % level of significance to determine whether the population variances for the number of days until resale are different in Houston than in Chicago. Assume the numbers of days resale houses are on the market are normally distributed.
Houston Chicago
132 126 118 56
138 94 85 69
131 161 113 67
127 133 81 54
99 119 94 137
126 88 93
134
X2 Goodness –of-Fit TestThe x2 goodness-of-fit test compares
Expected (theoretical) frequencies
Of categories from a population distribution
To the observed (actual) frequencies
From a distribution to determine whether
There is a difference between what was
Expected and what was observed.
This test can be used to determine whether the observed arrivals at teller windows at a bank are Poisson distributed, as might be expected. Also in the paper industry, manufactures can use the Chi-square good-ness-of-fit test to determine whether the demand for paper follows a uniform distribution throughout the year.
X2 Goodness-of-Fit Test
Formula to be used to compute the chi-square goodness-of-fit test.
data sample thefrom estimated parameters ofnumber =
categories ofnumber
valuesexpected offrequency
valuesobserved offrequency :
- 1 - = df
2
2
c
k
f
fwhere
ck
f
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The Chi-square Distribution
Let be a positive integer. Then a r.v. X is said to have a chi-squared distribution with parameter if the pdf of X is
The parameter is called the number of degrees of freedom of X.
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0 x 0
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2(2
12
12
2
xexx
Properties of the Chi-square distribution
(1) Distribution is a continuous probability distribution.
(2) The exact shape of the distribution depends upon the number of degrees of freedom . For different values of , we shall have different shapes of the distribution. In general when is small, the shape of the curve is skewed to the right and as gets larger, the distribution becomes more and more symmetrical and can be approximated by the normal distribution.
(3) The mean of the chi-square distribution is and variance 2 .
(4) The sum of independent chi-square variates is also a chi-square variate .
(5) The chi-square distribution is the sum of the squares of k independent random variables and there fore can never be less than zero
v v
v v
v
Exercise 12.7.The general manager of a major league baseball team believes the ages of purchasers of game tickets are normally distributed. The following data represent the distribution of ages for a sample of observed purchasers of major league baseball game tickets. Use the chi-square good ness of fit test to determine whether this distribution is
significantly different from the normal distribution. Assume .
Age of purchaser Frequency
10-under 20 16
20-under 30 44
30-under 40 61
40-under 50 56
50-under 60 35
60-under 70 19
05.0
Exercise 12.8. The Springfield Emergency Medical Service keeps records of emergency telephone calls. A study of 150 five-minutes time intervals resulted in the distribution of number of calls as follows. For example, during 18 of the five minutes intervals, no calls occurred. Use the chi square goodness of fit test and
to determine whether this distribution is Poisson.
Number of calls
(per 5-minute interval) Frequency
0 18
1 28
2 47
3 21
4 16
5 11
6 or more 9
56
01.0
Exercise 12.2 Use the following data and to determine whether the observed frequencies represent a uniform distribution.
Category F0
1 19
2 17
3 14
4 18
5 19
6 21
7 18
8 18
01.0
Chi-square test of independence
Goodness of fit test cannot be used to analyze two variables simultaneously.
The chi-square test of independence can be used to analyze the frequencies of two variables with multiple categories to determine whether the two variables are independent.
Example: A market researcher might want to determine whether the type of soft drink preferred by a consumer is independent of the consumer’s age. Financial investors might want to determine whether the type of preferred stock investment is independent of the region where the investor resides.
Suppose a business researcher is interested in determining whether geographic region is independent of type of financial investment. On a questionnaire, the following two questions might be used to measure geographic region and type of financial investment.
•In which region of the country do you reside?A. Northeast B. Midwest C. South D. West
•Which type of financial investment are you most likely to make today?
E. Stocks F. Bonds G. Treasury bills
Type of financialInvestment
E F GA O13 nA
Geographic B nB
Region C nC
D nD
nE nF nG N
Contingency Table
Type of Financial Investment
E F GA e12 nA
Geographic B nB
Region C nC
D nD
nE nF nG N
Contingency Table
2 Test of Independence: Formulas
Expected Frequency
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i row of total the
column thej
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Exercise: Suppose a business researcher wants to determine whether type of gasoline preferred is independent of a persons income. She takes a random survey of gasoline purchasers, asking them one question about the gasoline preference and the second question about the income. The respondent is to check whether he or she prefers (1) regular gasoline, (2) premium gasoline, or (3) extra premium gasoline. The respondent is also check his or her income brackets as being (1) less than $30000, (2) $30,000 to $49,999, (3) $50,000 to $99,999 or (4) more than $100,000. Using alpha=0.01, use the chi-square test of independence to determine whether type of gasoline preferred is independent of the income level.
Solution: Step-1
The hypothesis is
Step-2. The appropriate statistical test is
Step-3. Alpha is 0.01
Step-4. Here there are four rows (r=4) and three columns (c=3). The degrees of freedom are (4-1)(3-1)=6.
income. oft independennot is gasoline of Type
income oft independen is gasoline of Type 0
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ff 2
02
812.166,01.2
Step-5. Observed frequencies
Type of Gasoline
Income Regular PremiumExtra
PremiumLess than $30,000 85 16 6 107
$30,000 to $49,999 102 27 13 142$50,000 to $99,000 36 22 15 73At least $100,000 15 23 25 63
238 88 59 385
Expected Frequency:
Type of Gasoline
Income Regular PremiumExtra
PremiumLess than $30,000 (66.15) (24.46) (16.40)
85 16 6 107$30,000 to $49,999 (87.78) (32.46) (21.76)
102 27 13 142$50,000 to $99,000 (45.13) (16.69) (11.19)
36 22 15 73At least $100,000 (38.95) (14.40) (9.65)
15 23 25 63238 88 59 385
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12
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107 238
38566 15
107 88
38524 46
107 59
38516 40
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.
2 Calculation
2
2
88 6615 16 24 46 6 16 40
102 87 78 27 32 46 13 2176
36 4513 22 16 69 15 1119
15 38 95 23 14 40 25 9 65
66 15 24 46 16 40
87 78 32 46 21 76
4513 16 69 1119
38 95 14 40 9 6570 78
o ef ff e
2 2 2
2 2 2
2 2 2
2 2 2
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . ..
Conclusion :
0.01
df = 6
16.812
Non rejectionregion
.Hreject ,812.1678.70 o2 Cal
Exercise 12.13. Use the following contingency table and the chi-square test of independence to determine whether social class is independent of number of children in a family.
Social Class
Number of Children
Lower Middle Upper
0 7 18 6 31
1 9 38 23 70
2 or 3 34 97 58 189
3 47 31 30 108
97 184 117 398
H0 : Social Class is independent of Number of Children
Ha : Social Class is not independent of Number of Children
e11 = = 7.56 e31 = = 46.06
e12 = = 14.3 e31 = = 87.38
e13 = = 9.11 e31 = = 55.56
e21 = = 17.06 e31 = = 26.32
e22 = = 32.36 e31 = = 49.93
e23 = = 20.58 e31 = = 31.75
Social Class
Number of
Children
Lower Middle Upper
0(7.56)
7(14.33)
18(9.11)
631
1(17.06)
9(32.36)
38(20.58)
2370
2 or 3(46.06)
34(87.38)
97(55.56)
58189
3(26.32)
47(49.93)
31(31.75)
30108
97 184 117 398
Analysis of Variance
Experimental Design
• A plan and a structure to test hypotheses in which the researcher controls or manipulates one or more variables.
Independent Variable
• Treatment variable is one that the experimenter controls or modifies in the experiment.
• Classification variable is a characteristic of the experimental subjects that was present prior to the experiment, and is not a result of the experimenter’s manipulations or control.
Example : Suppose a manufacturing organization produces a valve that is specified to have an opening of 6.37 centimeters. Quality controllers within the company might decide to test to determine how the openings for produced valves vary among four different machines on three different shifts.
Here the independent variable is the type of machine and work shift. These are also classification variables as these are existed prior to the study.
Levels or Classifications are the subcategories of the independent variable used by the researcher in the experimental design.
In the valve experiment four levels or classification of machines with in the independent variable machine type are used : Machine1, Machine 2, Machine 3, and Machine 4
Dependent Variable
The response to the different levels of the independent variables. In the valve experiment the dependent variable is the size of the opening of the valve.
Analysis of Variance : experimental designs are analyzed statistically by a group of techniques known as Analysis of Variance.
Example : Suppose the measurements for the opening of 24 valves randomly selected from an assemble line are given below.
6.26 6.19 6.33 6.26 6.50
6.19 6.44 6.22 6.54 6.23
6.29 6.40 6.23 6.29 6.58
6.27 6.38 6.58 6.31 6.34
6.21 6.19 6.36 6.56
Analysis of Variance : Assumptions
Observations are drawn from normally distributed populations.
Observations represent random samples from the populations
Variances of the populations are equal.
One-way ANOVA : Computational Procedure
If k samples are being analyzed, the following hypotheses are being tested in a one-way ANOYA.
Ha : At least one of the means is different from the others
kH ....: 3210
MSE
MSCF
If FFc, reject Ho
If F Fc, do not reject Ho
One-Way ANOVA : Sums of Squares Definitions
Total sum of squares = error sum of squares + between sum of squares
SST = SSC+SSE
11
1
2
11
22
11
n
ijij
c
j
c
jjjji
n
i
c
j
xxxxnxx
valueindividualx
levelorgrouptreatmentaofmeanx
meangrandX
leveltreatmentgivenainnsobservatioofnumbern
levelstreatmentofnumberC
leveltreatmentaj
leveltreatmentaofmemberParticularWhere
ij
j
j
:
One-Way ANOVA : Computational Formulas
MSE
MSCF
df
SSEMSE
df
SSCMSC
NdfxxSST
CNdfxjxSSE
cdfxxnSSC
E
c
T
nj
iij
c
j
nj
iEji
c
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cj
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1
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2
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valueindividualx
meanColumnx
meangrandX
leveltreatmentgivenainnsobservatioofnumbern
levelstreatmentofnumberC
leveltreatmentaj
leveltreatmentaofmemberParticularaWhere
ij
j
j
:
The F-distribution is named in honour of R.A. Fisher who first studied it in 1924. This distribution is usually defined in terms of the ratio of the variances of two normally distributed populations. The quantity is distributed as F distribution with and
degrees of freedom.
If then the statistic follows F distribution with
degrees of freedom
Properties of F-distribution
1) The F-distribution is positively skewed and its skew ness decreases with increase in
2) The value of F must always be positive or zero since variances are squares and can never assume negative values. Its value will always lie between o and infinity.
122 nv
111 nv
22
12
222
2
12
12
s
s
22
12
s
sF
21 vandv
21 vandv
3) The shape of the F distribution depends upon the number of degrees of freedom
4) The mean and variance of the F-distribution are
2
22
21
2122
22
2
42
22
2
vforvvv
vvvVariance
vforv
vMean
2
4
In ANOVA situation the F value is a ratio of the treatment variance to the error variance.
Exercise : 11.11. A milk company has four machines that fill gallons 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 quarts) for 19 jugs of milks filled by the different machines. Use
The test the hypothesis.
Machine 1 Machine 2 Machine 3 Machine 4
4.05 3.99 3.97 4.00
4.01 1.02 3.98 4.02
4.02 4.01 3.97 3.97
4.04 3.99 3.95 4.01
4.00 4.00
4.00
Ex-11.13 A management consulting company presents a three day seminar on project management to various clients. The seminar is basically the same each time it is given. However, sometimes it is presented to high level managers, the seminar facilitators believe evaluations of the seminar may vary with the audience. Suppose the following data are some randomly selected evaluations scores from different levels of managers who attended the seminar. The ratings are on a scale from 1to 10, with 10 being the highest. Use a one way ANOVA to determine whether there is a significant difference in the evaluations according to managers level. Assume alpha=0.05. Discuss the business implications of your findings.
High Level Mid Level Low Level
____________________________________________________
7 8 5
7 9 6
8 8 5
7 10 7
9 9 4
10 8
8
_____________________________________________________
A Factorial Design (Two Way ANOVA)
Some experiments are designed so that two or more treatments (independent variable) are explored simultaneously. Such experimental designs are referred to as factorial designs.
Factorial Designs with two treatments.
Cells
.
.
.
.
.
.
.
.
.
.
.
.
Column Treatment
RowTreatment
.
.
.
.
.
Here there are two independent variables (two treatments) and there is an intersection of each level of each treatment. These intersections are referred to as cells.
Examples: The natural gas industry can design an experiment to study usage rates and how they are affected by temperature and precipitation. Theorizing that the outside temperature and type of precipitation make a difference in natural gas usage, industry researchers can gather usage measurements for a given community over a variety of temperature and precipitation conditions. At the same time they can make an effort to determine whether certain types of precipitation combined with certain temperature levels, affect usage rates differently than other combination of temperature and precipitation (interaction effect)
Statistically testing the Factorial Design
Analysis of variance is used to analyze data gathered from factorial designs. For factorial design with two factors (Independent variables) a two way ANOVA is used to test hypothesis statistically.
Two-Way ANOVA: Hypotheses: Following hypothesis is tested for two way ANOVA
Row Effects: H : Row Means are all equal.
H : At least one row mean is different from the others.
Columns Effects: H : Column Means are all equal.
H : At least one column mean is different from the others.
Interaction Effects: H : The interaction effects are zero.
H : There is an interaction effect.
o
a
o
a
o
a
Formulas for computing two-way ANOVA
mean grand = X
meancolumn = X
mean row = X
mean cell = X
nobservatio individual = X
member cell =k
levelatment column tre = j
level treatmentrow = i
s treatmentrow ofnumber = R
atmentscolumn tre ofnumber = C
cellper nsobservatio ofnumber =n
:
1
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11)(
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j
i
ij
ijk
1 1 1
2
1 1 1
2
1 1
2
1
2
1
2
where
nRC
SSEMSE
MSE
MSIF
CR
SSIMSI
MSE
MSCF
C
SSCMSC
MSE
MSRF
R
SSRMSR
NdfXXSST
nRCdfXXSSE
CRdfXXXXnSSI
CdfXXnRSSC
RdfXXnCSSR
I
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c
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r
n
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k
I
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Exercise 11.42. Children are gradually believed to have considerable influence over their parents in the purchase of certain items, particularly food and beverage items. To study this notion further, a study is conducted in which parents are asked to report how many food and beverage items purchased by the family per week are purchased because of the influence of their children. Because the age of the child may have an effect on the study, parents are asked to focus on one particular child in the family, for the week, and to report the age of the child. Four age categories are selected for the children: 4-5 years, 6-7 years, 8-9 years, and 10-12 years. Also because the number of children in the family might make a difference, three different sizes of families are chosen for the study: families with one child, families with two children, and families with three or more children. Suppose the following data represent the reported number of child-influenced buying incidents per week. Use the data to compute a two way ANOVA. Let alpha=0.05.
Number of Children in Family
_________________________________________________
1 2 3 or more
__________________________________________________
4-5 2 1 1
4 2 1
Age of 6-7 5 3 2
Child 4 1 1
(years) 8-9 8 4 2
6 5 3
10-12 7 3 4
8 5 3
____________________________________________________
Correlation and Regression Correlation is a measure of degree of relatedness of variables.
• In a bi-variate distribution we may be interested to find out if there is any correlation between the two variables under study. If the change in one variable affects the change in the other variables the variables are said to be correlated.
• IF the two variables deviate in the same direction, correlation is said to be positive.
Examples: Height and weight of a group of persons. Income and expenditure.
• If the variables constantly deviate in the opposite direction, the correlation is said to be negative.
• Examples: Price and demand of a commodity. Volume and pressure of a perfect gas.
Perfect correlation: Correlation is said to be perfect if the deviation in one variable is followed by a corresponding proportional deviation in the other.
Methods of ascertaining whether two variables are correlated or not.
I. Scatter diagram method.
II. II. Karl Pearson’s Coeff of Correlation.
III. III. Spearman’s rank correlation coefficient.
Scatter diagram method.
Plot the points on the graph.
• If all the points lie on a straight line falling from lower left hand corner to the upper right hand corner, correlation is said to be perfectly positive.
• If the pts lying on a straight line rising from the upper left hand corner to the lower right hand corner, correlation is said to be perfectly negative.
• If the plotted points fall in a narrow band there will be high degree of correlation between the variables.
• If the points are widely scattered in the diagram it indicated a very low degree of relationship.
Given the following pair of values
Capital employed 1 2 3 4 5 6 7 8 9 11 12
(Crores in Rs.)
Profit (Lakhs in Rs.) 3 5 4 7 9 8 10 11 12 14
(a) Make a scatter diagram
(b) Do you think that there is any correlation between profits and capital employee? Is it positive? Is it high or low?
Karl Pearson Coefficient of Correlation.
As a measure of intensity or degree of linear relationship between two variables, Karl Pearson a British biometrician, developed a formula called correlation coefficient
Correlation coefficient between two random variables X and Y, usually denoted by r (X, Y) or , is numerical measure of linear relationship between them and is defined as
1),(1
)()(
)()(),(),(
22
YXr
YYXX
YYXXyxCovyxr
yx
Exercise. Given below are the monthly income and their net savings of a sample of 10 supervisory staff belonging to a firm. Calculate the correlation coefficient.
Employee Number
1 2 3 4 5 6 7 8 9 10
Monthly Income
780 360 980 250 750 820 900 620 650 390
Net Savings 84 51 91 60 68 62 86 58 53 47
No X Y x y xy
1 780 84 130 18 16900 324 2340
2 360 51 -290 -15 84100 225 4350
3 980 91 330 25 108900 625 8250
4 250 60 -400 -6 160000 36 2400
5 750 68 100 2 10000 4 200
6 820 62 170 -4 28900 16 -680
7 900 86 250 20 62500 400 500
8 620 58 -30 -8 900 64 240
9 650 53 0 -13 0 169 0
10 390 47 -260 -19 67600 361 4640
Total 6500 660 0 0 539800 2224 27040
)(78.0)2224()539800(
2704022
approxyx
xyr
Rank Correlation Coefficient.
This method of finding out the covariability or the lack of it between two variables was developed by the British psychologist Charles Edward Spearman in 1904.
The measure is specially useful when the quantitative measure of certain factors (such as in the evaluation of leadership ability or the judgment of female beauty) cannot be fixed, but the individuals in a group can be arranged in order thereby obtaining for each individuals a number indicating his or her rank in a group.
Spearman’s Rank Correlation Coefficient
Where R = rank coeff of correlation
D = Difference of ranks between paired items in two series.
N = Number of data points.
)1(
61
2
2
NN
DR
Exercise. Two house wives, Geeta and Rita, asked to expenses their preferences for different kinds of detergents, gave the following replies:
To what extent the preferences of these two ladies go together.
Detergent Geeta Rita
A 4 4
B 2 1
C 1 2
D 3 3
E 7 8
F 8 7
G 6 5
H 5 6
I 9 9
J 10 10
In order to find out how far the preferences for different kinds of detergents go together, we will calculate the rank correlation coeff.
Detergent Rank by Geeta
Rank by Rita
1 4 4 0
B 2 1 1
C 1 2 1
D 3 3 0
E 7 8 1
F 8 7 1
G 6 5 1
H 5 6 1
I 9 9 0
J 10 10 0
N = 10
Thus the preference of these two ladies agree very closely as far as their opinion on detergent is concerned.
960.0990
661
61
3
2
1
NN
DR
Equal ranks or tie in ranks.
In some cases it may be necessary to assign equal rank to two or more individual or entries.
A correction is adjusted to the formula.
Where is the number of item whose rank is same in group 1 And so on.
NN
mmmmD
R
3
2321
31
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12
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Exercise. An examination of eight applicants for a clerical post was taken by a firm. From the marks obtained by the applicants in the Accountancy and Statistics papers, compute the rank correlation coefficient.
Applicant A B C D E F G H
Marks in Accountancy
15 20 28 12 40 60 20 80
Marks in Statistics
40 30 50 30 20 10 30 60
Applicant Marks inAccountancy
Ranks assigned Marks in Statistics
Rank Assigned by
A 15 2 30 6 16.00
B 20 3.5 30 4 0.25
C 28 5 50 7 4.00
D 12 5 30 4 9.00
E 40 1 20 2 16.00
F 60 6 10 1 36.00
G 20 7 30 4 0.25
H 80 3.5 60 8 0.00
N = 8 8
There is no correlation between the marks obtained in the two subjects.
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12
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1
2
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2
1
RThus
mm
NN
mmmmD
R
Merits and limitations of rank method.
1. This method is is simpler to understand and easier to apply compared to the Karl Person’s method. The answers obtained by this method and the Karl Pearson’s method is the same provided the no value is repeated that is all the items are different.
2. Where the data are of qualitative nature like honesty, efficiency, intelligence etc this method can be used with great advantage.
3. This method is the only method when we are given the ranks
and not the actual data.
4. Even where actual data are given, rank method can be applied for ascertaining rough degree of correlation.
Limitations: calculations becomes tedious when the number of data points exceed 30.
Regression Analysis.
Regression analysis is the process of constructing a mathematical model or function that can be used to predict or determine one variable by another variable.
• bivariate (two variables) linear regression
- the most elementary regression model
- dependent variable, the variable to be predicted, usually called Y
- independent variable, the predictor or explanatory variable, usually called X
• Deterministic Regression Model
Y = 0 + 1X
• 0 and 1 are population parameters
• 0 and 1 are estimated by sample statistics b0 and b1
Equation of the Sample Regression Line
YofvaluepredictedtheY
slopesampletheb
erceptsamplethebwhere
xbbY
1
0
10
int:
n
Xb
n
YXbYb
SSxx
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n
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Exercise 13.3: A corporation owns several companies. The strategic planner for the corporation believes dollar spent on advertising can to some extent be a predictor of total sales dollars. As an aid to the long term planning, she gathers the following sales and advertising information from several of the companies for 2002 ($millions).
Develop the equn of the simple regression line to predict sales from advertising expenditures using these data.
Advertisement Sales
12.5 148
3.7 55
21.6 338
60.0 994
37.6 541
6.1 89
16.8 126
41.2 379
Standard Error of Estimate
Residuals: Each difference between the actual y values and the predicted y values is the error of the regression line at a given point, and is referred to as residuals.
Standard Error of Estimate: To examine the error of the model we need to calculate he standard error of estimate, which provides a single measurement of the regression error.
SSE = Sum of the squares of the error =
Standard error of estimate =
Short cut formula =
yy
2yy
2
n
SSESe
xybybY 102
Exercise 13.25: Determine the equation of the regression line to predict annual sales of a company from the yearly stock market volume of shares sold in a recent year. Compute the standard error of the estimate for this model. Does volume of shares sold appear to be a good predictor of a company’s sales? Why or why not?
Company Annual sales ($ billions)
Annual Volume (millions of shares)
Merk 10.5 728.6
Phillip Morris 48.1 497.9
IBM 64.8 439.1
Eastman Kodak 20.1 377.9
Bristol-Myers Squibb 11.4 375.5
General Motors 123.8 363.8
Ford Motors 89.0 276.3
Coefficient of Determination.
The error sum of squares SSE can be interpreted as measure of how much variation in y is left unexplained by the model- That is how much cannot be attributed to a linear relationship.
A quantitative measure of the total amount of variation in observed y values is given by the total sum of squares.
The ratio SSE/SST is the proportion of the total variation that cannot be explained by the sample linear regression model, and 1-SSE/SST (a number between 0 and 1) is the proportion of the observed y variation explained by the model.
The Coefficient of Determination :
SST
SSEr 12
The higher the value of r2, the more successful is the simple linear regression model in explaining y variation.
Exercise 13.31. The Conference Board produces a Consumer Confidence Index (CCI) that reflects people’s feelings about general business conditions, employment opportunities, and their own income prospects. Some researchers may feel that consumer confidence is a function of the median household income. Shown here are the CCIs for 9 years and the median household incomes for the same 9 years published by the U.S. Census Bureau. Determine the equation of the regression line to predict the CCI from the median household income. Compute the standard error of the estimate for this model. Compute the value of r2. Explain the meaning.
CCI Median Household Income ($1000)
116.8 37.415
91.5 36.770
68.5 35.501
61.6 35.047
65.9 34.700
90.6 34.942
100.0 35.887
104.6 36.306
125.4 37.005
Prediction Interval to Estimate Y for a given value of X
n
XXSSxx
XofvalueparticularaXWhere
SSxx
XX
nSfY e
na
22
0
2
0
2,2
:
11
13.39. A Specialist in hospital administration stated that the number of FTEs (full time employees) in a hospital can be estimated by counting the number of beds in the hospital (a common measure of hospital size). A health care business researcher decided to develop a regression model in an attempt to predict the number of FTEs of a hospital by the number of beds. She surveyed 12 hospitals and obtained the following data. The data are presented in sequence, according to the number of beds.
Number of beds FTEs Number of beds FTEs
23 69 50 138
29 95 54 178
29 102 64 156
35 119 66 184
42 126 76 176
46 125 78 225
Construct a 90% interval for a single value of y. Use x = 100.
Exercise 13.40: Construct a 98% prediction interval for a single value of y for problem 13.3. Use x = 20
