By NEAS - 7/1/2024 4:11:02 PM
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
|
|