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final_exam_ec315.pdf
PART I. HYPOTHESIS TESTING PROBLEM 1 A certain brand of fluorescent light tube was advertised as having an effective life span before burning out of 4000 hours. A random sample of 84 bulbs was burned out with a mean illumination life span of 1870 hours and with a sample standard deviation of 90
hours. Construct a 95 confidence interval based on this sample and be sure to interpret this interval. Answer Since population standard deviation is unknown, t distribution can be used construct the confidence interval.
The 95% confidence interval is given by / 2, 1 / 2, 1
, n n
S S X t X t
n n
Details
Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation 90
Sample Mean 1870
Sample Size 84
Confidence Level 95%
Intermediate Calculations
Standard Error of the Mean 9.819805061
Degrees of Freedom 83
t Value 1.988959743
Interval Half Width 19.53119695
Confidence Interval
Interval Lower Limit 1850.47
Interval Upper Limit 1889.53
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PROBLEM 2 Given the following data from two independent data sets, conduct a one-tail hypothesis test to determine if the means are statistically equal using alpha=0.05. Do NOT do a confidence interval. n1 = 35 n2 = 30 xbar1= 32 xbar2 = 25 s1=7 s2 = 6 Answer H0:µ1=µ2 H1: µ1>µ2
Test statistics used is 1 1
1 2 1 2 2
1 2
~ n n
X X n n t t
S n n
where
2 2
1 1 2 2
1 2
( 1) ( 1)
2
n S n S S
n n
Decision rule : Reject the null hypothesis, if the calculated value of test statistic is greater than the critical value. Details
t Test for Differences in Two Means
Data
Hypothesized Difference 0
Level of Significance 0.05
Population 1 Sample Sample Size 35
Sample Mean 32
Sample Standard Deviation 7
Population 2 Sample Sample Size 30
Sample Mean 25
Sample Standard Deviation 6
Intermediate Calculations
Population 1 Sample Degrees of Freedom 34
Population 2 Sample Degrees of Freedom 29
Total Degrees of Freedom 63
Pooled Variance 43.01587
Difference in Sample Means 7
t Test Statistic 4.289648
Upper-Tail Test Upper Critical Value 1.669402
p-Value 3.14E-05
Reject the null hypothesis
Conclusion: Reject the null hypothesis. The sample provides enough evidence to support the claim that means are different.
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PROBLEM 3. A test was conducted to determine whether gender of a display model affected the likelihood that consumers would prefer a new product. A survey of consumers at a trade show which used a female spokesperson determined that 120 of 300 customers preferred the product while 92 of 280 customers preferred the product when it was shown by a female spokesperson. Do the samples provide sufficient evidence to indicate that the gender of the salesperson affect the likelihood of the product being favorably regarded by consumers? Evaluate with a two-tail, alpha =.01 test. Do NOT do a confidence interval. Answer H0: There no significant gender wise difference in the proportion customers who preferred the product. H1: There significant gender wise difference in the proportion customers who preferred the product.
The test Statistic used is Z test 1 2
1 2
1 1 (1 )
P P Z
P P n n
where p= 1 1 2 2
1 2
n p n p
n n
Decision rule : Reject the null hypothesis, if the calculated value of test statistic is greater than the critical value. Details
Z Test for Differences in Two Proportions
Data
Hypothesized Difference 0
Level of Significance 0.01
Group 1 Male
Number of Successes 120
Sample Size 300
Group 2 Female
Number of Successes 92
Sample Size 280
Intermediate Calculations
Group 1 Proportion 0.4
Group 2 Proportion 0.328571429
Difference in Two Proportions 0.071428571
Average Proportion 0.365517241
Z Test Statistic 1.784981685
Two-Tail Test Lower Critical Value -2.575829304
Upper Critical Value 2.575829304
p-Value 0.074264288
Do not reject the null hypothesis
Conclusion: Fails to reject the null hypothesis. The sample does not provide enough evidence to support the claim that there significant gender wise difference in the proportion customers who preferred the product.
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PROBLEM 4 Assuming that the population variances are equal for Male and Female GPA’s, test the following sample data to see if Male and Female PhD candidate GPA’s (Means) are equal. Conduct a two-tail hypothesis test at α =.01 to determine whether the sample means are different. Do NOT do a confidence interval. Answer H0: There is no significant difference in the mean GPA of males and Females
H1: There is significant difference in the mean GPA of males and Females.
Test Statistic used is independent sample t test.
1 1
1 2 1 2 2
1 2
~ n n
X X n n t t
S n n
where
2 2
1 1 2 2
1 2
( 1) ( 1)
2
n S n S S
n n
Decision rule: Reject the null hypotheses, if the calculated value of test statistic is greater
than the critical value.
Details
t Test for Differences in Two Means
Data
Hypothesized Difference 0
Level of Significance 0.05
Population 1 Sample Sample Size 12
Sample Mean 2.8
Sample Standard Deviation 0.25
Population 2 Sample Sample Size 13
Sample Mean 4.95
Sample Standard Deviation 0.8
Intermediate Calculations
Population 1 Sample Degrees of Freedom 11
Population 2 Sample Degrees of Freedom 12
Total Degrees of Freedom 23
Pooled Variance 0.363804
Male GPA’s Female GPA’s
Sample Size 12 13
Sample Mean 2.8 4.95
Sample Standard Dev .25 .8
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Difference in Sample Means -2.15
t Test Statistic -8.90424
Two-Tail Test
Lower Critical Value -2.80734 Upper Critical Value 2.807336 p-Value 0.0000
Reject the null hypothesis
Conclusion: Reject the null hypotheses. The sample provides enough evidence to support the claim that there is significant difference in the mean GP A score among the males and females.
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PART II REGRESSION ANALYSIS Problem 5 You wish to run the regression model (less Intercept and coefficients) shown below:
VOTE = URBAN + INCOME + EDUCATE Given the Excel spreadsheet below for annual data from1970 to 2006 (with the data for row 5 thru row 35 not shown), complete all necessary entries in the Excel Regression Window shown below the data.
Regression Input OK
Input Y Range:
Cancel
Input X Range:
Help
Labels Constant is Zero
Confidence Level: 95 %
Output options
Output Range:
New Worksheet Ply:
New Workbook
Residuals
Residuals Residual Plots
Standardized Residuals Line Fit Plots
Normal Probability
Normal Probability Plots
A B C D E
1 YEAR VOTE URBAN INCOME EDUCATE
2 1970 49.0 62.0 7488 4.3
3 1971 58.3 65.2 7635 8.3
4 1972 45.2 75.0 7879 4.5
36 2004 50.1 92.1 15321 4.9
37 2005 67.7 94.0 15643 4.7
38 2006 54.2 95.6 16001 5.1
x
X
X
X
A1:A38
B1:E38
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PROBLEM 6. Use the following regression output to determine the following: A real estate investor has devised a model to estimate home prices in a new suburban development. Data for a random sample of 100 homes were gathered on the selling price of the home ($ thousands), the home size (square feet), the lot size (thousands of square feet), and the number of bedrooms. The following multiple regression output was generated:
Regression Statistics
Multiple R 0.8647
R Square 0.7222
Adjusted R Square 0.6888
Standard Error 16.0389
Observations 100
Coefficients Standard
Error t Stat P-value
Intercept -24.888 38.3735 -0.7021 0.2154
X1 (Square Feet) 0.2323 0.0184 9.3122 0.0000
X2 (Lot Size) 11.2589 1.7120 4.3256 0.0001
X3 (Bedrooms) 15.2356 6.8905 3.2158 0.1589
a. Why is the coefficient for BEDROOMS a positive number? The selling price increase when the number of rooms increases. Thus the relationship is positive.
b. Which is the most statistically significant variable? What evidence shows this? Most statistically significant variable is one with least p value. Here most statistically significant variable is Square feet.
c. Which is the least statistically significant variable? What evidence shows this? Least statistically significant variable is one with high p value. Here least statistically significant variable is bedrooms
d. For a 0.05 level of significance, should any variable be dropped from this model? Why or why not? The variable bed rooms can be dropped from the model as the p value is greater than 0.05.
e. Interpret the value of R squared? How does this value from the adjusted R squared?
The R 2 gives the model adequacy. Here R
2 suggest that 72.22% variability can be explained by
the model.
Adjusted R 2 is a modification of R
2 that adjusts for the number of explanatory terms in a model.
Unlike R 2 , the adjusted R
2 increases only if the new term improves the model more than would
be expected by chance.
f. Predict the sales price of a 1134-square-foot home with a lot size of 15,400 square feet and 2 bedrooms. Selling Price =-24.888+0.02323*1134+11.2589*15400+15.2356*2=173419
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PART III SPECIFIC KNOWLEDGE SHORT-ANSWER QUESTIONS. Problem 7 Define Autocorrelation in the following terms:
a. In what type of regression is it likely to occur?
Regressions involving time series data
b. What is bad about autocorrelation in a regression? The standard error of the estimates will high.
c. What method is used to determine if it exists? (Think of statistical test to be used) Durbin Watson Statistic is used determine auto correlation in a regression.
d. If found in a regression how is it eliminated?
Appropriate transformations can be adopted to eliminate auto correlation.
Problem 8 Define Multicollinearity in the following terms:
a) In what type of regression is it likely to occur?
Multicollinearity occurs in multiple regressions when two or more independent
variables are highly correlated.
b) Why is multicollinearity in a regression a difficulty to be resolved?
Multicollinearity in Regression Models is an unacceptably high level of
intercorrelation among the independents, such that the effects of the independents
cannot be separated. Under multicollinearity, estimates are unbiased but assessments
of the relative strength of the explanatory variables and their joint effect are unreliable.
c) How can multicollinearity be determined in a regression?
Multicollinearity refers to excessive correlation of the predictor variables. When
correlation is excessive (some use the rule of thumb of r > 0.90), standard errors of the
b and beta coefficients become large, making it difficult or impossible to assess the
relative importance of the predictor variables. The measures Tolerance and VIF are
commonly used to measure multicollinearity. Tolerance is 1 - R2 for the regression of
that independent variable on all the other independents, ignoring the dependent.
There will be as many tolerance coefficients as there are independents. The higher the
inter-correlation of the independents, the more the tolerance will approach zero. As a
rule of thumb, if tolerance is less than .20, a problem with multicollinearity is
indicated. When tolerance is close to 0 there is high multicollinearity of that variable
with other independents and the b and beta coefficients will be unstable. The more the
multicollinearity, the lower the tolerance, the more the standard error of the regression
coefficients.
d) If multicollinearity is found in a regression, how is it eliminated?
Multicollinearity occurs because two (or more) variables are related – they measure
essentially the same thing. If one of the variables doesn’t seem logically essential to
your model, removing it may reduce or eliminate multicollinearity.
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