 Independent variables in a regression equation should NOT be mutually exclusive.
 Pvalue is the smallest level of α at which to reject Hο.
 The usual objective of regression analysis is to predict/estimate the value of one variable when the value of another variable is known.
 Correlation analysis is concerned with measuring the strength of the relationship between two variables.
 In the least squares model, the explained sum of squares is always smaller than the regression sum of squares.
 The sample correlation coefficient and the sample slope will always have the same sign.
 An important relationship in regression analysis is = .
 If in a regression analysis the explained sum of squares is 75 and the unexplained sum of square is 25, r2 = 0.33.
 When small values of Y tend to be paired with small values of X, the relationship between X and Y is said to be inverse.
 The probability that the test statistic will fall in the critical region, given that H0 is true, represents the probability of making a type II error.
 When we reject a true null hypothesis, we commit a Type I error.
Part B: Multiple Choice (12–21)
 A computer statistical package has included the following quantities in its output: SST = 50, SSR = 35, and SSE = 15. How much of the variation in y is explained by the regression equation?
 49% 70% c. 35% d. 15%
 In testing the significance of b_{1}, the null hypothesis is generally that
 β = b_{1} β 0 c. β = 0 d. β = r
 Testing whether the slope of the population regression line could be zero is equivalent to testing whether the population could be zero.
 standard error of estimate yintercept
 prediction interval coefficient of correlation
 A multiple regression equation includes 4 independent variables, and the coefficient of multiple determination is 0.64. How much of the variation in y is explained by the regression equation?
 80% 16% c. 32% d. 64%
 A multiple regression analysis results in the following values for the sumofsquares terms: SST = 50.0, SSR = 35.0, and SSE = 15.0. The coefficient of multiple determination will be
 = 0.35 = 0.30 c. = 0.70 d. = 0.50
 In testing the overall significance of a multiple regression equation in which there are three independent variables, the null hypothesis is
 In a multiple regression analysis involving 25 data points and 4 independent variables, the sumofsquares terms are calculated as SSR = 120, SSE = 80, and SST = 200. In testing the overall significance of the regression equation, the calculated value of the test statistic will be
 F = 1.5 F = 5.5
 F = 2.5 F = 7.5
 For a set of 15 data points, a computer statistical package has found the multiple regression equation to be = 23 + 20+ 5 + 25 and has listed the tratio for testing the significance of each partial regression coefficient. Using the 0.05 level in testing whether = 20 differs significantly from zero, the critical t values will be
 t = 1.960 and t= +1.960
 t = 2.132 and t = +2.132
 t = 2.201 and t = +2.201
 t = 1.796 and t = +1.796
 Computer analyses typically provide a pValue for each partial regression coefficient. In the case of , this is the probability that
 = 0
 =
 the absolute value of could be this large if = 0
 the absolute value of could be this large if 1
 In the multiple regression equation, = 20,000 + 0.05+ 4500 , is the estimated household income, is the amount of life insurance held by the head of the household, and is a dummy variable ( = 1 if the family owns mutual funds, 0 if it doesn’t). The interpretation of = 4500 is that
 owing mutual funds increases the estimated income by $4500
 the average value of a mutual funds portfolio is $4500
 45% of the persons in the sample own mutual funds
 the sample size must have been at least n = 4500
Part C: Please fill in the blank, circle your decision or answer the following questions (2224).
 State whether you would reject or fail to reject the null hypothesis in each of the following cases (twotailed): Make a decision.
P = 0.12 ; 
Decision 
Reject / Fail to reject 
P = 0.03 ; 
Decision 
Reject / Fail to reject 
P = 0.001 ; 
Decision 
Reject / Fail to reject 
 Given the following, complete the ANOVA table and make the correct inference. Using Fvalue to make a decision.
Source 
SS 
df 
MS 
F 
Treatments 

3 


Error 
88.8 



Total 
435 
19 



ANSWER 
a) What is the hypothesis being tested in this problem? 
H_{0}=The treatment are all equal. H_{a}=at least one of them is not equal or all of them are not equal 
b) In the above ANOVA table, is the factor significant at the 5% level? 

c) What is the number of observations? 

 State whether should be accepted or rejected for , given the following; (fill in the blank and circle your decision)
 a)= 2.34; k=3, n=14
Computed F 
Critical F 
Decision 
2.34 
________ 
Reject / Fail to reject 
 b)= 2.52; k=5, n=25
Computed F 
Critical F 
Decision 
2.52 
________ 
Reject / Fail to reject 
 c)= 4.29; k=4, n=28
Computed F 
Critical F 
Decision 
4.29 
________ 
Reject / Fail to reject 
Part D: Must show all your work step by step in order to receive the full credit; Excel is not allowed. (2535)
 Consider the following hypothesis test.
H_{o}: µ = 17
H_{a}: µ ≠ 17
A sample of 25 gives a sample mean of 14.2 and sample variance of 25.
a) 
At 5% should the null be rejected? 
b) 
Compute the value of the test statistic 
c) 
What is the pvalue? 
d) 
What is your conclusion? Explain. 
 Consider the following hypothesis test
H_{o}: µ ≥ 10
H_{a}: µ < 10
A sample of 50 provides a sample mean of 9.46 and sample variance of 4.
a) 
At 5% should the null be rejected? 
b) 
Compute the value of the test statistic 




c) 
What is the pvalue? 
d) 
What is your conclusion? 
 Use problem 13 on page 928 to answer the following questions.
A bath soap manufacturing process is designed to produce a mean of 120 bars of soap per batch. Quantities over or under the standard are undesirable. A sample of ten batches shows the following number bars of soap.
108 118 120 122 119 113 124 122 120 123
Using a 0.05 level of significance, test to see whether the sample results indicate that the manufacturing process is functioning properly.
a) What is the sample mean

b) What is the sample standard deviation

c) Use Z or T test? And why?

d) What is your hypothesis test

e) At α = 0.05, what is the rejection rule?

f) Compute the value of the test statistic.

g) What is the pvalue?

h) What is your conclusion?

 Fill the below table and answer the given questions
Months on job (x) 
Monthly sales (y) thousands of dollars 
X^{2} 
Y^{2} 
XY 





1 
0.80 








2 
2.40 








4 
7.00 








5 
3.70 








8 
11.30 








9 
12.00 








12 
15.00 








Total 
52.2 








a)Find 
b)Find

c)Write the equation and interpret 
d) Compute R^{2 }and how is it different from adjusted R^{2}.

e) Compute the estimated variance of the regression. 
f) Find

g) Compute the estimated variance of 
h) Compute the standard error of

 Please fill in the computer printout and answer the following questions. Given that
SUMMARY OUTPUT 





Regression Statistics 





Multiple R 
0.9037 




R Square 
______ 




Adjusted R Square 
______ 




Standard Error 
______ 




Observations 
5 




ANOVA 






df 
SS 
MS 
F 
Significance F 
Regression 
1 
4.9 
_____ 
_____ 
0.03535 
Residual 
3 
1.1 
_____ 


Total 
_____ 
_____ 




Coefficients 
Standard 
t Stat 
Pvalue 
Lower95% 
Upper 95% 
Lower 95.0% 
Upper 95.0% 
Intercept 
0.1 
______ 
0.15746 
0.88488 
2.12112 
1.92112 
2.12112 
1.92112 
X1 
0.7 
______ 
3.65563 
0.03535 
0.09061 
1.30939 
______ 
______ 
a) What percent of the variation is explained by the regression equation?

b) What is the standard error of regression?

c) What is the critical value of the Fstatistic? 
d) What sample size is used in the print out 
 The following regression equation was obtained using the five independent variables.Given that
a) What percent of the variation is explained by the regression equation? 
b) What is the standard error of regression? 2 
c) Write the estimated equation. 
d) What is the critical value of the Fstatistic? 
e) What sample size is used in the print out? 
f) What is the variance of the slope coefficient of income?

g) Assuming that you are using a twotailed test make a decision using the computed Pvalue. 
 A large hotel purchased 200 new color televisions several months ago: 80 of one brand and 60 of each of two other brands. Records were kept for each set as to how many service calls were required, resulting in the table that follows.
Number of Service Calls 
TV Brand 
Total 

Sony 
Toshiba 
Sanyo 

None 
8 
15 
18 
41 
One 
30 
55 
12 
97 
Two or more 
22 
10 
30 
62 
Total 
60 
80 
60 
200 
Assume the TV sets are random samples of their brands. With 5% risk of Type I error, test for
an association between TV brand and the number of service calls.
Is the value significant at 5% level of significant? Write the conclusion for this question.
 A metropolitan bus system sampler’s rider counts on one of its express commuter routes for week. Use the following data to establish whether ridership is evenly balanced by day of the week. Let = 0.05.
Day 
Monday 
Tuesday 
Wednesday 
Thursday 
Friday 
Rider Count 
10 
34 
21 
57 
44 
Is the value significant at 5% level of significant? Write the conclusion for this question
 Explain the major difference between and ANOVA in terms of type of data (no calculation required)
 Explain why a 95% confidence interval estimate for the mean value of y at a particular x is narrower than a 95% confidence interval for an individual y value at the same value of x
 If , n=11, andwhat is ? (Single Regression model)
 You are given the following information from fitting a multiple regression with three variables to 30 sample data points:
PART A: True or False
1) F
2) T
3) T
4) T
5) F
6) T
7) T
8) F
9) F
10) F
11) T
PART B: Multiple Choice
12) Option B
13) Option C
14) Option D
15) Option D
16) Option C
17) Option A
18) Option D
19) Option C
20) Option A
21) Option A
PART C: Fill in the blanks
22) a) Fail to reject
 b) Reject
 c) Reject
23) a) Fail to Reject
 b) Fail to Reject
 c) Reject
24 a) H_{0}=The treatment are all equal.
H_{a}=at least one of them is not equal or all of them are not equal
 b) Yes, Factor is significant
 c) 20
Question 25
H_{o}: µ = 17
H_{a}: µ ≠ 17
Sample size = 25
Sample mean = 14.2
Sample variance = 25
Standard deviation = sqrt(25) = 5
a) 
At 5% should the null be rejected? 
b) 
Compute the value of the test statistic 

(a) H_{0} will be rejected when the p value is higher than significance level (5%). 

(b) The test statistics

c) 
What is the pvalue? 
d) 
What is your conclusion? Explain. 

The p value (for 24 degree of freedom and 2.8 t value) = 0.0099


It can be seen that p value is lower than significance level (0.05) and hence, null hypothesis would be rejected and alternative hypothesis would be accepted. Hence, it can be said that hypothesized mean is not same as 17. 
Question 26
H_{o}: µ >= 10
H_{a}: µ < 10
Sample size = 50
Sample mean = 9.46
Sample variance = 4
Standard deviation = sqrt(4) = 2
a) 
At 5% should the null be rejected? 
b) 
Compute the value of the test statistic 

(a) H_{0} will be rejected when the p value is higher than significance leel (5%). 

(b) The test statistics 
c) 
What is the pvalue? 
d) 
What is your conclusion? 

The p value (for 49 degree of freedom and 1.91 t value) = 0.031 

(c) It can be seen that p value is lower than significance level (0.05) and hence, null hypothesis would be rejected and alternative hypothesis would be accepted. Hence, it can be said that population mean is less than 10. 
Question 27
What is the sample mean Sample mean = 108+118+120+122+119+113+124+122+120+123 = 118.9

What is the sample standard deviation Standard deviation = sqrt(218.9/9) =4.93 Standard deviation = 4.93 
Use Z or T test? And why? Population standard deviation is unknown and also, the sample size is lower than 30 and hence, t test would be taken into consideration. 
What is your hypothesis test Null hypothesis H_{0}: µ = 120 Alternative hypothesis H_{a}: µ ≠ 120 
At α = 0.05, what is the rejection rule? At significance level 0.05, the null hypothesis would be rejected when p value is lower than 0.05. 
Compute the value of the test statistic.

What is the pvalue? The p value (for (101) = 9 degree of freedom and –0.70 t value) = 0.5016

What is your conclusion? It can be seen that p value is higher than significance level (0.05) and hence, null hypothesis would not be rejected and alternative hypothesis would not be accepted. Hence, it cannot be said that mean number of bars is difference from 120. 
Question 28
a)Find 
b)Find 
c)Write the equation and interpret y = 1.3152 x0.2462 
d) Compute R^{2 }and how is it different from adjusted R^{2}. 
e) Compute the estimated variance of the regression. 
f) Find 
g) Compute the estimated variance of Variance of b1 = MSE Sxx Sxx = sum(x x bar)^2 = 94.86 Variance of b1 = MSE Sxx = 2.487/94.86 = 0.026 
h) Compute the standard error of Standard error of = sqrt(2.487/94.86) = 0.162 
Question 29
a) What percent of the variation is explained by the regression equation? R^2 = (49)/6.0 = 0.8167 Required percentage = 81.67% 
b) What is the standard error of regression?
Standard error = sqrt((10.816)^2 * 0.1915) = 0.1874 
c) What is the critical value of the Fstatistic? F stat = 13.36 
d) What sample size is used in the print out?
Sample size = 5 
Question 30
a) What percent of the variation is explained by the regression equation? 99.4% of variation is explained by the regression equation. 
b) What is the standard error of regression? 2 Standard error of regression is 1.507. 
c) Write the estimated equation. Sales = 19.7 0.00063outlets +1.74cars +0.410income + 2.04age 0.034 bosses 
d) What is the critical value of the Fstatistic? Critical value is 6.256 based on the given significance level coupledwith degrees of freedom. 
e) What sample size is used in the print out? Sample size = 10 
f) What is the variance of the slope coefficient of income?
Variance = √ SE * n^{4} = (0.04385)^{0.5}*10^{4} = 2,094.04 
g) Assuming that you are using a twotailed test make a decision using the computed Pvalue. The null hypothesis would be rejected since the p value is zero which is lesser than the significance level. As a result, the alternative hypothesis would be accepted. 
Question 31
Null and alternative hypothesis
H_{o}: TV brand and number of service calls are independent.
H_{a}: TV brand and number of service calls are dependent.
= .5033 + 0.1195 + 2.6415 + 0.0278 + 6.7639 + 10.0485 + 0.6215 + 8.8323 +6.987 = 37.54
Degree of freedom = (31) (31) =4
The p value for 37.54 chi square and 4 degree of freedom = 0.00
Significance level = 5%
It can be seen from the above that p value is lower than level of significance and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Hence, it can be concluded that TV brand and number of service calls are dependent.
Question 32
Null and alternative hypothesis
H_{o}: Ridership is equally balanced.
H_{a}: Ridership is not equally balanced.
Expected frequency = (10+34+21+57+44)/5 = 33.2
= 41.2892
Degree of freedom = 51 = 4
The p value for 37.54 chi square and 4 degree of freedom = 0.00
Significance level = 5%
It can be seen from the above that p value is lower than level of significance and hence, sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Hence, it can be concluded that ridership is not equal on all days.
 b) The major difference between chisquare and ANOVA is that the chisquare is used when the data is of categorical type and nonnumerical in nature. On the contrary, ANOVA is used when the underlying data is numerical with the level of measurement being interval or ratio.
Question 33
When the estimation of mean value of y is done, then standard deviation with regards to sample mean is given by s/√n
Thus, the confidence interval is given by (Sample mean – margin of error, Sample mean + margin of error)
Thus, the width of the confidence interval is 2*margin of error = 2*critical value * s/√n
When the y’s individual value is estimated for the value of x which is same, then s would be standard deviation
Thus, the width of the confidence interval is 2*s
As a result, the comparison of the two confidence interval width clearly supports the assertion in the question.
Question 34
R^2 =0.95
N = 11
SST= 100
Now,
Question 35
a) Null and alternative hypothesis H_{o}: H_{a}: . Test stat (t value) = (25.20)/15 =1.68 Critical value of t = 2.05 It can be seen that t stat is lower than critical value of t and hence, cannot reject null hypothesis. Therefore, it can be concluded that slope is insignificant and can be assumed to be zero at 5% significance level. 
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