MS Module 5 t values and confidence intervals practice exam questions

(The attached PDF file has better formatting.)

A statistician estimates the population mean for a normal distribution from a sample of 6 points. The 99% confidence interval for the population mean is (0, 1.78)


Question 5.1: Critical t value

What is the critical t value for a 99% confidence interval from a sample of 6 points?

Answer 5.1: 4.032 (table look-up)


Question 5.2: Standard error of estimated mean

What is the standard deviation of the sample / the square root of the number of observations?

Answer 5.2: (1.78 – 0) / (2 × 4.032) = 0.2207

(width of confidence interval = 2 × t value × σ/√N)


Question 5.3: Standard deviation of sample

What is the standard deviation of the sample?

Answer 5.3: 0.2207 × 60.5 = 0.5406

(width of confidence interval = 2 × t value × σ/√N)


Question 5.4: Critical t value

What is the critical t value for a 95% confidence interval from a sample of 6 points?

Answer 5.4: 2.5706 (table look-up)


Question 5.5: Estimated mean

What is the estimated mean of the population?

Answer 5.5: (1.78 – 0) / 2 = 0.89

(mean = mid-point of confidence interval)


Question 5.6: Confidence interval

What is the 95% confidence interval for the population mean?

Answer 5.6: The confidence interval is 0.89 ± 0.2207 × 2.5706:

●    lower bound: 0.89 – 0.2207 × 2.5706 = 0.323
●    upper bound: 0.89 + 0.2207 × 2.5706 = 1.457