Q2. 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 that the variability of industry sales explains about 95.2% of the variability of company sales.

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

Q3. What is the predicted value for sales of Company XYZ given industry sales of $3,500?

A)   $994.88.

B)   $900.00.

C)   $883.72.

Q4. 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.

答案和详解如下：

Q2. 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 that the variability of industry sales explains about 95.2% of the variability of company sales.

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

Correct answer is B)

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.

Q3. What is the predicted value for sales of Company XYZ given industry sales of $3,500?

A)   $994.88.

B)   $900.00.

C)   $883.72.

Correct answer is C)

The regression equation is Y = (−94.88) + 0.2796 × X = −94.88 + 0.2796 × (3,500) = 883.72.

Q4. 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.

Correct answer is A)

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

The upper limit for a 95% confidence interval = Ŷ + t_cs_f = 827.8 + 3.182 × 15.5028 = 827.8 + 49.33 = 877.13 (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