Question

Problem 10.12. A course can be taken for credit either by attending lectures at fixed times and days, or by taking online sessions that can be done a the student's own pace and at those times/days that the student chooses. The course coordinator wants to determine if these two ways of taking the course resulted in a significant difference in achievement as measured by the final exam for the course. The table below gives the scores on an examination with 45 possible points for one group of ni = 9 students who took the course online, and a second group of n2 = 9 students who took the course with conventional lectures. Online 32 37 35 28 41 44 35 31 34 Classroom 35 31 29 25 34 40 27 32 31 (a) First compute the sample means X and Y, as well as the sample variances Sį and S, of the two groups. (b) Note that Sị and S3 are quite close. For this reason, you decide that the two populations have the same variance, and you decide to employ the pooled variance method, described at the end of this homework assignment, to compute the 95% confidence interval for the difference of the two means for the two populations. Does the number zero belong to such interval? (If so, this suggests that there no significant difference in the grade for students who take the course online vs. the students who take it with conventional lectures. This idea is strictly connected with the notion of hypothesis testing, which we will be exploring this week.)

Answer 1

SOLUTION-

1.) WE USE MINITAB-16 FOR COMPUTING SAMPLE MEAN AND SD FOR THE TWO GIVEN SAMPLES.

STEPS- ENTER THE DATA IN SEPERATE COLUMNS> STAT> BASIC STATISTICS> DISPLAY DESCRIPTIVE STATISTICS> SELECT BOTH THE VARIABLES> UNDER 'STATISTICS', SELECT 'MEAN' AND 'STANDARD DEVIATION'> OK

VARIABLE	SAMPLE SIZE	SAMPLE MEAN	SAMPLE SD
ONLINE	$n_1=9$	$\bar{x}_1=35.22$	$1 = 4.94
CLASSROOM	$n_2=9$	$\bar{x}_2=31.56$	$s_2=4.48$

2.) POOLED STANDARD DEVIATION IS DEFINED AS,

(nı - 1)s + (n2 - 1)s ni+n2 - 2

SO, $S_p=\sqrt{\frac{(9-1)*4.94^{2}+(9-1)*4.48^{2}}{9+9-2}}=4.71$

ALSO, CRITICAL VALUE = $t_{\alpha /2,n_1+n_2-2}= t_{0.025,16}=2.120$

MARGIN OF ERROR(E) = $t*S_p*\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

= $2.120*4.71*\sqrt{\frac{1}{9}+\frac{1}{9}}= 4.71$

HENCE, THE 95% CONFIDENCE INTERVAL FOR DIFFERENCE OF MEAN IS,

  $(\bar{x}_1-\bar{x}_2)\pm E$

= $(35.22-31.56)\pm 4.71$

=

YES, THE 95% CONFIDENCE INTERVAL CONTAINS ZERO.

**** IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.