Question

A professor has a teaching assistant record the amount of time (in minutes) that a sample of 16 students engaged in an active discussion. The assistant observed 8 students in a class who used a slide show presentation and 8 students in a class who did not use a slide show presentation.

With Microsoft PowerPoint	Without Microsoft PowerPoint
19	18
6	5
13	10
11	8
23	9
15	21
14	12
7	4

Use the normal approximation for the Mann-Whitney U test to analyze the data above. (Round your answer to two decimal places.)
z =  

State whether to retain or reject the null hypothesis. (Assume alpha equal to 0.05.)

Retain the null hypothesis. Reject the null hypothesis.   

 

Answer 1

 

Let us first distributed the data in ascending order and then give them ranks .the table below shows the same

with microsoft power point	Rank 1	without microsoft power point	Rank 2
		4	1
		5	2
6	3
7	4
		8	5
		9	6
		10	7
11	8
		12	9
13	10
14	11
15	12
		18	13
19	14
		21	15
23	16

In Mann-Whitney U test  

= - / 2 /

= + ( + 1 ) / 2 -

= + ( + 1 ) / 2 -

= ( , )

=   / 2

=   + ( + + 1 /12)

Use the normal approximation for the Mann-Whitney U test to analyze the data above. (Round your answer to two decimal places.)

Here,

=8

=8

= sum of Ranks 1 = 3+4+8+10+11+12+14+16 = 78

= sum of Ranks 2 = 1+2+5+6+7+9+13+15 = 58

= + ( + 1 ) / 2 -

= 8*8 + 8 ( 8 + 1 ) / 2 - 78​​​​​​​

= 22

= + ( + 1 ) / 2 -

= 8*8 + 8 ( 8 + 1 ) / 2 - 58

= 42

= ( , )

= ( 22 , 42 )

= 22

=   / 2

= 8*8 / 2

= 32

=   + ( + + 1 /12)

= 8*8 + ( 8 + 8 + 1 /12)   

= 90.666

= 22 - 8*8 / 2 /

= 22 - 32/

= −10 / 9.5218

= −1.050221597

= −1.05 (Round your answer to two decimal places.)

State whether to retain or reject the null hypothesis. (Assume alpha equal to 0.05.)Retain the null hypothesis. Reject the null hypothesis

Assume = 0.05

At

= −1.05 and   = 0.05

p-value = 0 .146859. (for one - tailed)

= 0.293718. (for two-tailed)

hence,

p-value > so Reject the null hypothesis