MS Module 17 Confidence interval and prediction interval practice exam questions

(The attached PDF file has better formatting.)

[The practice problems in the 24 modules explain the statistical procedures; the practice exam questions in this thread shows what you will be asked on the final exam.]

A linear regression analysis relates Y to X.

●    The X values are {1, 2, …, 7}
●    The least squares estimator for β1 is a linear function of the Y values =  ciYi
●    The error sum of squares (SSE) is 25


Question 17.1:

What is , the average X value?

Answer 17.1: (1 + 7) / 2 = 4


Question 17.2: Sxx

What is Sxx, the sum of squared residuals for the X values?

Answer 17.2: (1 – 4)2 + (2 – 4)2 + (3 – 4)2 + (4 – 4)2 + (5 – 4)2 + (6 – 4)2 + (7 – 4)2 = 28


Question 17.3: Least squares estimate of σ2

What is s2, the estimate of σ2?

Answer 17.3: 25 / (7 – 2) = 5

(least squares estimate of σ2 = SSE / (number of observations – 2) )


Question 17.4: Least squares estimate of σ

What is s, the estimate of σ?

Answer 17.4: 50.5 = 2.2361

(standard deviation = square root of variance)


Question 17.5: Standard deviation of the least squares estimate of β1

What is the standard deviation of the least squares estimate of β1?

Answer 17.5: 2.2361 / 280.5 = 0.4226

(standard deviation = square root of variance)


Question 17.6: t value for 90% two-sided confidence interval

What is the t value for a 90% two-sided confidence interval?

Answer 17.6: 2.015

(Table look-up)


Question 17.7: Width of confidence interval

What is the width of the 90% confidence interval at x = 2?

Answer 17.7: 2 × 2.2361 × 2.015 × (1 / 7 + (2 – 4)2 / 28)0.5 = 4.8168

(width of the confidence interval is 2 × tα/2,n-2 × s × (1/n + (x* – )2 / Sxx )½ )


Question 17.8: Width of prediction interval

What is the width of the 90% prediction interval at x = 2?

Answer 17.8: 2 × 2.2361 × 2.015 × (1 + 1 / 7 + (2 – 4)2 / 28)0.5 = 10.2181

(width of the confidence interval is 2 × tα/2,n-2 × s × (1 + 1/n + (x* – )2 / Sxx )½ )