MS Module 2: Confidence intervals – practice problems

(The attached PDF file has better formatting.)

Exercise 2.1: Confidence interval

A sample from a normal distribution has summary statistics:

●    n = 50
●    xi = 991
●    xi2 = 20,635

A.    What is the estimated μ, the mean of the normal distribution?
B.    What is the estimated σ, the standard deviation of the normal distribution?
C.    What is the 95% confidence interval for μ?

Part A: The estimated μ = 991 / 50 = 19.82.

Part B: The estimated σ = ( (20,635 – 9912/50) / (50 – 1) )0.5 = 4.5026

Part C: The 95% confidence interval is

●    Lower bound: 19.82 – 1.96 × 4.5025 / 500.5 = 18.5720
●    Upper bound: 19.82 + 1.96 × 4.5025 / 500.5 = 21.0680

Exercise 2.2: Confidence intervals

A statistician forms confidence intervals for the mean of a normally distributed population from a sample of 80 observations.

●    The upper bound of the 95% confidence interval is 5.
●    The lower bound of the 90% confidence interval is 1.

A.    What is the estimated standard deviation of the population?
B.    What is the estimated mean of the population?

Part A: Let μ be the estimated mean and σ be the estimated standard deviation.

●    The (1–z) confidence interval for the mean is (μ – zα/2 × σ, μ + zα/2 × σ).
●    The zα/2 values are 1.96 for a 95% confidence interval and 1.645 for a 90% confidence interval.

We have two equations in two unknowns:

μ + 1.96 × σ/√n = 5
μ – 1.645 × σ/√n = 1

The first equation minus the second equation gives

σ/√n = (5 – 1) / (1.96 + 1.645) = 1.1096, so σ = 1.1096 × 800.5 = 9.9246

Part B: μ = 5 – 1.96 × 1.1096 = 2.8252

Exercise 2.3: μ and σ

A statistician estimates confidence intervals from a sample of N observations for the mean (μ) of a normal distribution with a known variance σ2.

●    The upper bound of the 95% confidence interval is 5.
●    The lower bound of the 90% confidence interval is 1.

A.    What is the zα/2 for the 95% confidence interval?
B.    What is the zα/2 for the 90% confidence interval?
C.    What is σ/√N, the standard deviation of the sample mean?
D.    What is the estimated mean ()?
E.    If N = 8, what is σ, the standard deviation of the normal distribution?

Part A: For the 95% confidence interval, zα/2 = z0.025 = 1.959964 (table look-up or spreadsheet function).

Part B: For the 90% confidence interval, zα/2 = z0.05 = 1.644854 (table look-up or spreadsheet function).

Part C: We have two equations in two unknowns: and σ/√N

●    (5 – ) = 1.959964 × σ / √N
●    ( – 1) = 1.644854 × σ / √N

Adding the two equations gives

(5 – 1) = (1.959964 + 1.644854) × σ / √N ➾

σ / √N = (5 – 1) / (1.959964 + 1.644854) = 1.109626

Part D: = 1 + 1.644854 × σ / √N = 1 + 1.644854 × 1.109626 = 2.825173

Alternatively, = 5 – 1.959964 × σ / √N = 5 – 1.959964 × 1.109626 = 2.825173

Part E: If N = 8, σ = 1.109626 × 80.5 = 3.138496