By NEAS - 8/7/2018 4:57:33 PM
MS Module 4 t values and two confidence intervals practice exam questions
A statistician estimates the population mean for a normal distribution from a sample of 6 points.
● The upper bound of the 99% confidence interval for the population mean is 5.78.
● The lower bound of the 95% confidence interval for the population mean is 1.4.
Question 4.1: Critical t value
What is the critical t value for a 99% confidence interval from a sample of 6 points?
Answer 4.1: 4.0321 (table look-up)
Question 4.2: Critical t value
What is the critical t value for a 95% confidence interval from a sample of 6 points?
Answer 4.2: 2.5706 (table look-up)
Question 4.3: Standard error of estimated mean
What is the standard error of the estimated mean of the population?
Answer 4.3: (5.78 – 1.4) / (4.0321 + 2.5706) = 0.6634
(standard error of the estimated mean of the population = (upper bound of first confidence interval – lower bound of second confidence interval) / (critical t value of first confidence interval + critical t value of second confidence interval)
Question 4.4: Standard deviation of the sample
What is the standard deviation of the sample?
Answer 4.4: 0.6634 × 60.5 = 1.625
(standard deviation of the sample = standard error of the estimated mean of the population × square root of the number of observations in the sample)
Question 4.5: Estimated mean of the population
What is the estimated mean of the population?
Answer 4.5: (two formulas:)
● 5.78 – 4.0321 × 0.6634 = 3.105
● 1.4 + 2.5706 × 0.6634 = 3.105
(estimated mean of the population = upper bound of the confidence interval – critical t value × standard error of the estimated mean OR lower bound of the confidence interval + critical t value × standard error of the estimated mean)
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