Question

ATH241 23 In a sam standard deviation of 39.2. a.) What is the best point estimate of the mean number of hours it takes students to meet online course ple of 42 students enrolled in a hybrid course (one that contains both online and classroom ). the mean number of hours it took students to meet the online course objectives was 71 8 with a 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 93% confidence level? (round answer to the earest hundredth) d) What is the 93% confidence interval estimate of the mean number of hours it takes students to meet online course objectives? (round values to three decimal places) e) Calculate the margin of error for a 93%, confidence interval. (round to three decimal places) 11:02 PM

Answer 1

a)

best point estimate=71.8

b)

degree of freedom=   DF=n-1= 42-1 = 41

c)

Level of Significance ,    α =    0.07
't value='   tα/2=   1.86    [Excel formula =t.inv(α/2,df) ]

d)

sample std dev ,    s =    39.2
Sample Size ,   n =    42
Sample Mean,    x̅ =   71.8

Standard Error , SE =   s/√n =   6.0487
margin of error ,   E=t*SE =   11.253

confidence interval is       
Interval Lower Limit=   x̅ - E =    60.547
Interval Upper Limit=   x̅ + E =    83.053

   So, confidence interval is (   60.547   < µ <    83.053   )

e)

margin of error ,   E=t*SE = 1.8605*6.0487 = 11.253