Question

In a sample of 20 students enrolled in a hybrid course one that contains both online and classroom instruction), the mean number of hours it took students to meet the online course objectives was 53.7 with a standard deviation of 12.1. a.) What is the best point estimate of the mean number of hours it takes students to meet online course objectives? b.) How many degrees of freedom will be used for the given sample size? c.) What is the positive critical value that corresponds to a 90% confidence level? (round answer to the nearest thousandth) d.) What is the 90% confidence interval estimate of the mean number of hours it takes students to meet online course objectives? (round values to two decimal places) < < e.) Calculate the margin of error for a 90% confidence interval. (round to two decimal places) E =

Answer 1

Given that Number of random samples = n = 20 Sample mean = X = 53.7 Sample standard deviation = s = 12.1 a) Sample mean = X = 53.7 Best point estimate = 53.7 b) Degrees of freedom = df = n-1 = 20 -1 = 19 :: Degrees of freedom = 19

c) Now we have to find the positive critical value corresponds to 90% confidence level. Step (1): =1-200 = 1 – 0.90 = 0.10 0.10 = 0.05 Step (2): 1 -0.05 = 0.95 Degrees of freedom = df = n - 1 = 20 – 1 = 19 Use excel formula: =T.INV(0.95,19), to get t = 1.729 .: positive critical value = t = 1.729

Now we have to find the margin of error. margin of error = t * in = 1.729 * 1273 = 4.68 :: margin of error = MOE = 4.68 d) Lower bound= Best estimate – margin of error = 53.7 – 4.68 = 49.02 Upper bound = Best estimate + margin of error = 53.7 + 4.68 = 58.38 : confidence interval = 49.02 <u < 58.38

note, that, option(d) is followed by option(e).