MS Module 5 t values and confidence intervals practice problems

(The attached PDF file has better formatting.)

Exercise 5.1: T test, one sample

A sample of N=15 observations from a normal distribution has

●    xi = 120 (sum of the N observations)
●    xi2 = 3,600 (sum of the squares of the N observations)

●    The null hypothesis is H0: the population mean μ0 = 4
●    This exercise shows the procedure for one-tailed and two-tailed tests, with either
    ○    The two-tailed alternative hypothesis is Ha: the population mean μ0 ≠ 4
    ○    The one-tailed (upper tailed) alternative hypothesis is Ha: the population mean μ0 > 4

A.    What is the sample mean of the N observations?
B.    What is the sample standard deviation of the N observations?
C.    What is the t value used to test the null hypothesis?
D.    What is the p value for the one-tailed alternative hypothesis?
E.    What is the p value for the two-tailed alternative hypothesis?

Part A: The sample mean is xi / N = 120 / 15 = 8.

Part B: The sample standard deviation = { [ xi2 – (xi)/N ] / (N – 1) }0.5 =

    ( (3,600 – 1202 / 15) / (15 – 1) )0.5 = 13.732131

Part C: The t value used to test the null hypothesis is ( – μ0) / (σ / N0.5) =

    (8 – 4) / (13.732131 /150.5) = 1.128152

Part D: The p value for the one-tailed alternative hypothesis = 0.1391 (table look-up or spreadsheet function)

Part E: The p value for the two-tailed alternative hypothesis = 0.2782 (table look-up or spreadsheet function)


Exercise 5.2: t values and confidence intervals

A statistician estimates the population mean for a normal distribution from a sample of 8 points. The 99% confidence interval for the population mean is (0, 2.500)

A.    What is the critical t value for a 99% confidence interval from a sample of 8 points?
B.    What is the standard deviation of the sample?
C.    What is the critical t value for a 95% confidence interval from a sample of 8 points?
D.    What is the 95% confidence interval for the population mean?

Part A: A sample with 8 points has 7 degrees of freedom. The critical t value for a 99% confidence interval with 7 degrees of freedom is 3.499 (table lookup).

Part B: The confidence interval is the estimate ± σ/√N × t value, so

●    the point estimate is (2.500 + 0) / 2 = 1.250
●    the width of the confidence interval is 2 × σ/√N × t value = 2.500 – 0 = 2.500 ➾
    ○    σ = 80.5 × 2.500 / (2 × 3.499) = 1.010

Part C: The critical t value for a 95% confidence interval with 7 degrees of freedom is 2.365 (table lookup).

Part D: The confidence interval is 1.250 ± (1.010 / 80.5) × 2.365:

●    lower bound: 1.250 – (1.010 / 80.5) × 2.365 = 0.405
●    upper bound: 1.250 + (1.010 / 80.5) × 2.365 = 2.095



Exercise 5.3: t values and two confidence intervals

A statistician estimates the population mean for a normal distribution from a sample of 8 points.

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

We use the following notation:

●     μ = the estimated population mean
●     σ = the standard deviation of the sample
●     N = the number of observations

A.    What is the critical t value for a two-sided 95% confidence interval for a sample of 8 points?
B.    What is the critical t value for a two-sided 90% confidence interval for a sample of 8 points?
C.    What is σ/√N?
D.    What is σ?
E.    What is μ?

Part A: A sample of 8 points has 8 – 1 = 7 degrees of freedom.

The critical t value for a two-sided 95% confidence interval for a sample of 8 points is 2.3646 (table look-up).

Part B: A sample of 8 points has 8 – 1 = 7 degrees of freedom.

The critical t value for a two-sided 90% confidence interval for a sample of 8 points is 1.8946 (table look-up).

Part C: Combine the upper half of the 95% confidence interval with the lower half of the 90% confidence interval to estimate σ/√N:

●    2.3646 × σ/√N = 5 – μ
●    1.8946 × σ/√N = μ – 1

adding: (2.3646 + 1.8946) × σ/√N = (5 – μ) + (μ – 1) = (5 – 1) ➾

σ/√N = (5 – 1) / (2.3646 + 1.8946) = 0.93914

Part D: σ = 0.93914 × 80.5 = 2.656

Part E: Solve for μ by either of the confidence intervals:

●    2.3646 × σ/√N = 5 – μ ➾ μ = 5 – 2.3646 × 0.93914 = 2.779
●    1.8946 × σ/√N = μ – 1 ➾ μ = 1 + 1.8946 × 0.93914 = 2.779