Question

A study was done to determine if students learn better in an online basic statistics class versus a traditional face-to-face (f2f) course. A random sample of 12 students in an online course and 12 students in a f2f course was taken. The scores (out of 100) on the final exam of both groups are shown below.Online class 60 100 95 80 67 88 76 86 90 95 91 100 f2f 55 70 90 81 67 89 34 76 94 88 100 67 93 76 89(1) What are the degrees of freedom for the pooled two-sample t procedure?(2) The standard error for the mean of the online group is _____.(3) The standard error for the mean of the f2f group is _____.

Answer 1

Answer:

1) 25

2) 3.507

3) 4.35

Step-by-step explanation:

Note: from the question, the number of F2F data is 15 not 12

1) for degree of freedom:

12 + 15 - 2 = 25

2) formula for standard error:

$S_e_r = A/(sqrt*N)$

But A is standard deviation. To find the standard deviation we use:

S.d = [sqrt*(E(xi - u)^2) / N]

Where xi = mean

u = individual data

Therefore

Mean = (60+100+95+80+67+88+76+86+90+95+91+100) / 12 =

1028/12 = 85.7

$(xi -u)^2 =$

$(85.7-60)^2 = 660.49$

$(85.7-100)^2= 204.49$

$(85.7-95)^2 = 86.49$

$(85.7 - 80)^2 = 32.49$

$(85.7 - 67)^2 = 349.69$

$(85.7-88)^2 = 5.29$

$(85.7-76)^2 = 94.09$

$(85.7 - 86)^2 = 0.09$

$(85.7 - 90)^2 = 18.49$

$(85.7-95)^2 = 86.49$

$(85.7-91)^2 = 28.09$

$(85.7-100)^2 = 204.9$

$E(xi- u)^2 = 1771.09$

$S.d = sqrt*[(E(xi-u)^2)/N]$

$= sqrt*(1771.09/12)$

$= 12.149$

So, standard error for online class=

$S_e_r = 12.149/√12 = 3.507$

3) We will use same method as in number two.

Mean= 1169/15 =77.93

S.d = √(4238.85/15) = 16.81

For standard error of F2F class:

$S_e_r = 16.81/sqrt*15 = 4.35$

Note: E represents summation here