JoeyDVivre

A sample of 200 monthly observations is used to run a simple linear regression:
Returns = b0 + b1Leverage + u.
The t-value for the regression coefficient of leverage is calculated as t = – 1.09.
A 5% level of significance is used to test whether leverage has a significant influence on returns.
The correct decision is to:

A) reject the null hypothesis and conclude that leverage does not significantly explain returns.

B) do not reject the null hypothesis and conclude that leverage does not significantly explain returns.

C) do not reject the null hypothesis and conclude that leverage significantly explains returns.

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Do not reject the null since |–1.09|<1.96(critical t-value).

JoeyDVivre

The most appropriate test statistic to test statistical significance of a regression slope coefficient with 45 observations and 2 independent variables is a:

A) one-tail t-statistic with 43 degrees of freedom.

B) two-tail t-statistic with 42 degrees of freedom.

C) one-tail t-statistic with 42 degrees of freedom.

---

df = n − k − 1 = 45 − 2 − 1

JoeyDVivre

Rebecca Anderson, CFA, has recently accepted a position as a financial analyst with Eagle Investments. She will be responsible for providing analytical data to Eagle’s portfolio manager for several industries. In addition, she will follow each of the major public corporations within each of those industries. As one of her first assignments, Anderson has been asked to provide a detailed report on one of Eagle’s current investments. She was given the following data on sales for Company XYZ, the maker of toilet tissue, as well as toilet tissue industry sales ($ millions). She has been asked to develop a model to aid in the prediction of future sales levels for Company XYZ. She proceeds by recalling some of the basic concepts of regression analysis she learned while she was preparing for the CFA exam.

Year	Industry Sales (X)	Company Sales (Y)	(X-X)2
1	$3,000	$750	84,100
2	$3,200	$800	8,100
3	$3,400	$850	12,100
4	$3,350	$825	3,600
5	$3,500	$900	44,100
Totals	$16,450	$4,125	152,000

Coefficient Estimates
Predictor	Coefficient	Stand. Error of the Coefficient	t-statistic
Intercept	-94.88	32.97	??
Slope (Industry Sales)	0.2796	0.0363	??

Analysis of Variance Table (ANOVA)
Source	df (Degrees of Freedom)	SS (Sum of Squares)	Mean Square (SS/df)	F-statistic
Regression	1 (# of independent variables)	11,899.50 (SSR)	11,899.50 (MSR)	59.45
Error	3 (n-2)	600.50 (SSE)	200.17 (MSE)
Total	4 (n-1)	12,500 (SS Total)

Abbreviated Two-tailed t-table
df	10%	5%
2	2.920	4.303
3	2.353	3.182
4	2.132	2.776

Standard error of forecast is 15.5028.Which of the following is the correct value of the correlation coefficient between industry sales and company sales?

A) 0.9062.

B) 0.9757.

C) 0.2192.

---

The R2 = (SST − SSE) / SST = (12,500 − 600.50) / 12,500 = 0.952
The correlation coefficient is √R2 in a simple linear regression, which is √0.952 = 0.9757. (Study Session 3, LOS 11.a)

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Which of the following reports the correct value and interpretation of the R2 for this regression? The R2 is:

A) 0.048, indicating that the variability of industry sales explains about 4.8% of the variability of company sales.

B) 0.952, indicating the variability of company sales explains about 95.2% of the variability of industry sales.

C) 0.952, indicating that the variability of industry sales explains about 95.2% of the variability of company sales.

---

The R2 = (SST − SSE) / SST = (12,500 − 600.50) / 12,500 = 0.952
The interpretation of this R2 is that 95.2% of the variation in company XYZ's sales is explained by the variation in tissue industry sales. (Study Session 3, LOS 11.a)

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What is the predicted value for sales of Company XYZ given industry sales of $3,500?

A) $994.88.

B) $883.72.

C) $900.00.

---

The regression equation is Y = (−94.88) + 0.2796 × X = −94.88 + 0.2796 × (3,500) = 883.72. (Study Session 3, LOS 11.h)

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What is the upper limit of a 95% confidence interval for the predicted value of company sales (Y) given industry sales of $3,300?

A) 877.13.

B) 827.87.

C) 318.42.

---

The predicted value is Ŷ = −94.88 + 0.2796 × 3,300 = 827.8.
The upper limit for a 95% confidence interval = Ŷ + tcsf = 827.8 + 3.182 × 15.5028 = 827.8 + 49.33 = 877.13.
The critical value of tc at 95% confidence and 3 degrees of freedom is 3.182.
(Study Session 3, LOS 11.h)

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What is the lower limit of a 95% confidence interval for the predicted value of company sales (Y) given industry sales of $3,300?

A) 778.47.

B) 827.80.

C) 1,337.06.

---

The predicted value is Ŷ = -94.88 + 0.2796 × 3,300 = 827.8.
The lower limit for a 95% confidence interval = Ŷ − tcsf = 827.8 − 3.182 × 15.5028 = 827.8 − 49.33 = 778.47.
The critical value of tc at 95% confidence and 3 degrees of freedom is 3.182.
(Study Session 3, LOS 11.h)

---

What is the t-statistic for the slope of the regression line?

A) 2.9600.

B) 7.7025.

C) 3.1820.

---

Tb = (b1hat − b1) / sb1 = (0.2796 − 0) / 0.0363 = 7.7025. (Study Session 3, LOS 11.g)

JoeyDVivre

Paul Frank is an analyst for the retail industry. He is examining the role of television viewing by teenagers on the sales of accessory stores. He gathered data and estimated the following regression of sales (in millions of dollars) on the number of hours watched by teenagers (TV, in hours per week):

Salest = 1.05 + 1.6 TVt

The predicted sales if television watching is 5 hours per week is:

A) $9.05 million.

B) $2.65 million.

C) $8.00 million.

---

The predicted sales are: Sales = $1.05 + [$1.6 (5)] = $1.05 + $8.00 = $9.05 million.

JoeyDVivre

Consider the regression results from the regression of Y against X for 50 observations:

Y = 5.0 - 1.5 X

The standard error of the estimate is 0.40 and the standard error of the coefficient is 0.45. The predicted value of Y if X is 10 is:

A) 10.

B) 20.

C) -10.

---

The predicted value of Y is: Y = 5.0 – [1.5 (10)] = 5.0 – 15 = -10

JoeyDVivre

Consider the regression results from the regression of Y against X for 50 observations:

Y = 5.0 + 1.5 X

The standard error of the coefficient is 0.50 and the standard error of the forecast is 0.52. The 95% confidence interval for the predicted value of Y if X is 10 is:

A) {19.480 < Y < 20.052}.

B) {18.980 < Y < 21.019}.

C) {18.954 < Y < 21.046}.

---

The predicted value of Y is: Y = 5.0 + [1.5 (10)] = 5.0 + 15 = 20. The confidence interval is 20 ± 2.011 (0.52) or {18.954 < Y < 21.046}.

JoeyDVivre

A variable Y is regressed against a single variable X across 24 observations. The value of the slope is 1.14, and the constant is 1.3. The mean value of X is 1.10, and the mean value of Y is 2.67. The standard deviation of the X variable is 1.10, and the standard deviation of the Y variable is 2.46. The sum of squared errors is 89.7. For an X value of 1.0, what is the 95% confidence interval for the Y value?

A) −1.68 to 6.56.

B) −1.83 to 6.72.

C) 0.59 to 4.30.

---

First the standard error of the estimate must be calculated—it is equal to the square root of the mean squared error, which is equal to the sum of squared errors divided by the number of observations minus 2 = (89.7 / 22)1/2 = 2.02. The variance of the prediction is equal to:


= 2.06
The prediction value is 1.3 + (1.0 × 1.14) = 2.44. The t-value for 22 degrees of freedom is 2.074. The endpoints of the interval are 2.44 ± 2.074 × 2.06 = −1.83 and 6.72.

JoeyDVivre

Given: Y = 2.83 + 1.5X
What is the predicted value of the dependent variable when the value of an independent variable equals 2?

A) 2.83

B) -0.55

C) 5.83

---

Y [/td][td]= 2.83 + (1.5)(2)
	= 2.83 + 3
	= 5.83

JoeyDVivre

A simple linear regression is run to quantify the relationship between the return on the common stocks of medium sized companies (Mid Caps) and the return on the S&P 500 Index, using the monthly return on Mid Cap stocks as the dependent variable and the monthly return on the S&P 500 as the independent variable. The results of the regression are shown below:

	Coefficient	Standard Error of Coefficient	t-Value
Intercept	1.71	2.950	0.58
S&P 500	1.52	0.130	11.69
R2 = 0.599

Use the regression statistics presented above and assume this historical relationship still holds in the future period. If the expected return on the S&P 500 over the next period were 11%, the expected return on Mid Cap stocks over the next period would be:

A) 20.3%.

B) 33.8%.

C) 18.4%.

---

Y = intercept + slope(X)

JoeyDVivre

A study of a sample of incomes (in thousands of dollars) of 35 individuals shows that income is related to age and years of education. The following table shows the regression results:

   
Coefficient
Standard Error
t-statistic
P-value

Intercept
5.65
1.27
4.44
0.01

Age
0.53
?
1.33
0.21

Years of Education
2.32
0.41
?
0.01

   

Anova
df
SS
MS
F

Regression
?
215.10
?
?

Error
?
115.10
?


Total
?
?




The standard error for the coefficient of age and t-statistic for years of education are:

A) 0.40; 5.66.

B) 0.53; 2.96.

C) 0.32; 1.65.





--------------------------------------------------------------------------------
standard error for the coefficient of age = coefficient / t-value = 0.53 / 1.33 = 0.40

t-statistic for the coefficient of education = coefficient / standard error = 2.32 / 0.41 = 5.66



--------------------------------------------------------------------------------
The mean square regression (MSR) is:
A) 6.72.

B) 107.55.

C) 102.10.





--------------------------------------------------------------------------------
df for Regression = k = 2

MSR = RSS / df = 215.10 / 2 = 107.55



--------------------------------------------------------------------------------
The mean square error (MSE) is:
A) 3.60.

B) 3.58.

C) 7.11.





--------------------------------------------------------------------------------

df for Error = n – k – 1 = 35 – 2 – 1 = 32

MSE = SSE / df = 115.10 / 32 = 3.60



--------------------------------------------------------------------------------
What is the R2 for the regression?
A) 76%.  

B) 62%.  

C) 65%.  





--------------------------------------------------------------------------------

SST = RSS + SSE

= 215.10 + 115.10

= 330.20

R2= RSS / SST = 215.10 / 330.20 = 0.65



--------------------------------------------------------------------------------
What is the predicted income of a 40-year-old person with 16 years of education?
A) $62,120.  

B) $74,890.

C) $63,970.  





--------------------------------------------------------------------------------

income  = 5.65 + 0.53 (age) + 2.32 (education)

            = 5.65 + 0.53 (40) + 2.32 (16)

            = 63.97 or $63,970



--------------------------------------------------------------------------------
What is the F-value?
A) 14.36.  

B) 1.88.

C) 29.88.





--------------------------------------------------------------------------------
F = MSR / MSE = 107.55 / 3.60 = 29.88