Q5. 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)   827.80.

B)   1,337.06.

C)   778.47.

Q6. What is the t-statistic for the slope of the regression line?

A)   2.9600.

B)   3.1820.

C)   7.7025.

Q7. A variable y is regressed against a single variable x across 28 observations. The value of the slope is 1.89, and the constant is 1.1. The mean value of x is 1.50, and the mean value of y is 3.94. The standard deviation of the x variable is 0.96, and the standard deviation of the y variable is 2.85. The regression sum of squares is 79.07, and the total sum of squares is 195.24. For an x value of 2.0, what is the 95% confidence interval for the y value?

A)   1.21 to 8.65.

B)   1.01 to 8.75.

C)   0.41 to 9.35.

Q8. 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):

Sales_t = 1.05 + 1.6 TV_t

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

A)    $2.65 million.

B)    $8.00 million.

C)    $9.05 million.

答案和详解如下：

Q5. 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)   827.80.

B)   1,337.06.

C)   778.47.

Correct answer is C)

The predicted value is Ŷ = -94.88 + 0.2796 × 3,300 = 827.8.

The lower limit for a 95% confidence interval = Ŷ − t_cs_f = 827.8 − 3.182 × 15.5028 = 827.8 − 49.33 = 778.47 (Interim calculations below).

The critical value of t_c at 95% confidence and 3 degrees of freedom is 3.182.

The standard error of the forecast, s_f²= s_e²[1 + 1/n + (X − X)²/(n − 1)s_x²], where

s_e² = MSE = 200.17
n = 5
X = 3,300
X = 16,450/5 = 3,290
s_x² = Σ(X_i − X)² / (n − 1) = 152,000 / 4 = 38,000

Substituting, s_f² = 200.17 × {1 + 1/5 + (3,300 − 3,290)² / [(5 − 1) × 38,000]} = 240.335691.

s_f = 15.5028

Q6. What is the t-statistic for the slope of the regression line?

A)   2.9600.

B)   3.1820.

C)   7.7025.

Correct answer is C)

Tb = (b₁hat − b₁) / s_b1 = (0.2796 − 0) / 0.0363 = 7.7025.

Q7. A variable y is regressed against a single variable x across 28 observations. The value of the slope is 1.89, and the constant is 1.1. The mean value of x is 1.50, and the mean value of y is 3.94. The standard deviation of the x variable is 0.96, and the standard deviation of the y variable is 2.85. The regression sum of squares is 79.07, and the total sum of squares is 195.24. For an x value of 2.0, what is the 95% confidence interval for the y value?

A)   1.21 to 8.65.

B)   1.01 to 8.75.

C)   0.41 to 9.35.

Correct answer is C)

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 residual sum of squares divided by the number of observations minus 2. The residual sum of squares is equal to the difference between the total sum of squares and the regression sum of squares, which is 195.24 − 79.07 = 116.17. The standard error of the estimate is equal to (116.17 / 26)^1/2 = 2.11. The standard deviation of the prediction is equal to the standard error of the estimate multiplied by {1 + 1/n + (X −X )²/[(n − 1)s_x²]}^1/2 = 2.11 × [1 + 1/28 + (2.0 − 1.5)²/(27 × 0.96²)]^1/2 = 2.17. The prediction value is 1.1 + (2.0 × 1.89) = 4.88. The t-value for 26 degrees of freedom is 2.056. The endpoints of the interval are 4.88 ± 2.056 × 2.17 = 0.41 and 9.35.

Q8. 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):

Sales_t = 1.05 + 1.6 TV_t

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

A)    $2.65 million.

B)    $8.00 million.

C)    $9.05 million.

Correct answer is C)

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

thx