## MS Module 2 Normal distribution practice exam questions

 Author Message NEAS Supreme Being         Group: Administrators Posts: 4.2K, Visits: 1.2K MS Module 2 Normal distribution practice exam questions.(The attached PDF file has better formatting.) A sample from a normal distribution has summary statistics:●    n = 10●    xi = 21●    xi2 = 159Question 2.1: Sample meanWhat is the mean of the sample?Answer 2.1: The mean of the sample is 21 / 10 = 2.1Question 2.2: Sample varianceWhat is the variance of the sample?Answer 2.2: The sum of squared deviations is 159 – 212 / 10 = 114.90 and the sample variance is     (159 – 212 / 10) / (10 – 1) = 12.767Question 2.3: Standard deviation What is the standard deviation of the sample?Answer 2.3: The standard deviation is the square root of the variance: ((159 – 212 / 10) / (10 – 1))0.5 = 3.573Question 2.4: Maximum likelihood estimate of the variance What is the maximum likelihood estimate of the variance? Answer 2.4: (159 – 212 / 10) / 10 = 11.490(The maximum likelihood estimate of the variance divides by N, not by (N-1). )Question 2.5: Maximum likelihood estimate of the standard deviationWhat is the maximum likelihood estimate of the standard deviation?Answer 2.5: ( (159 – 212 / 10) / 10 )0.5 = 3.390Question 2.6: Standard error of the sample meanWhat is the standard error of the sample mean?Answer 2.6: 3.573 / 100.5 = 1.130(Standard error of the mean = standard deviation of the sample / square root of the number of observations)Question 2.7: Confidence interval, lower boundWhat is the lower bound of the 90% two-sided confidence interval for the mean of the normal distribution?Answer 2.7: 2.100 – 1.645 × 1.130 = 0.241(For a confidence level of 90%, α = 10%, and zα/2 = 1.645 (table look-up). The lower bound of the two-sided confidence interval for the mean of the normal distribution = mean – zα/2 × standard error of the mean.)Question 2.8: Confidence interval, upper boundWhat is the upper bound of the 90% two-sided confidence interval for the mean of the normal distribution?Answer 2.8: 2.100 + 1.645 × 1.130 = 3.959 Attachments MS Module 2 Normal distribution practice exam questions.pdf (221 views, 43.00 KB)
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