Question

1.) Do employees of computer manufacturing perform better at work with music playing. The music was turned on during the working hours of a business with 45 employees. Their productivity level averaged 5.2 with a standard deviation of 2.4. On a different day the music was turned off and there were 40 workers. The workers' productivity level averaged 4.8 with a standard deviation of 1.2. What can we conclude at 90% level of confidence (compare CI for music on and music off)?

a. Employees perform better at work with music playing.

b.Employees performance doesn’t do anything with music playing.

c.Employees perform poorer at work with music playing.

d.None of the above.

Please Provide the Letter, I was told B is wrong

Answer 1

N1=45 ,M1= 5.2 AND SD.1= 2.4 AND N2=40, M2= 4.8 and SD2= 1.2

Diffrence of means = 5.2-4.8= 0.4

Standard error of mean= sqrt((2.4)^2/45+(1.2)^2/40)

= sqrt(0.128+0.036)

= sqrt(0.164)

=0.405

Degrees of freedom= 66

t critical value=1.668

90% CI DIFFERENCE OF MEANS

0.4 1.668*0.405

=0.4 0.676

= (-0.276, 1.076) IS 90% CONFIDENCE INTERVAL.

Since It does contain Zero so NOTSIGNIFICANT.

c.Employees perform poorer at work with music playing.