Question

I MUST CALCULATE THE HYPOTHESIS TEST ON BEFORE AND AFTER:

Ho: u1>u2

Ha: u1<u2

SAMPLE SIZE IS 58

BEFORE THE IMPROVEMENT: 40 EXPENSES

AFTER THE IMPROVEMENT: 18 EXPENSES

THE IMPROVEMENT PLAN: WAS 8 WEEKS (5 WEEKS (INITIAL) AND 3 WEEKS FOR THE CORRECTION).

BEFORE THE IMPROVEMENT:

•X=40

•N=58

•p=40/58=0.69

•Confidence Interval: After the improvement

•X=18

•N=58

•P=18/58=0.31

•Confidence Interval: (0.191, 0.429)

•95% confidence (1.96)that the interval contains the true population proportion.

•With a Margin of error: 0.12

•

Before the Improvement:

•

•Sample Standard Deviation, s 48.908185591827

•Variance (Sample Standard), s2 2392.0106178846

•Population Standard Deviation, σ 48.292963798441

•Variance (Population Standard), σ2 2332.2103524375

•Total Numbers, N 40

•Sum: 1484.89

•Mean (Average): 37.12225

•Standard Error of the Mean (SEx̄): 7.7330631348202

AFTER THE IMPROVEMENT:

•Sample Standard Deviation, s 17.276746092814

•Variance (Sample Standard), s2 298.48595555556

•Population Standard Deviation, σ 16.789979227776

•Variance (Population Standard), σ2 281.90340246914

•Total Numbers, N 18

•Sum: 277.82

•Mean (Average): 15.434444444444

•Standard Error of the Mean (SEx̄): 4.0721681063556

Answer 1

Solution :

Here we have to define the hypothesis test as,

Ho: u1>u2

Ha: u1<u2

Where u1 and u2 are means of population before and after respectively.

Population variance is known for us.

we have to use test of differences of population means.

Here we have to define test statistics as,

Zcal=

Let us consider, under null hypothesis,u1=u2

=(37.12225-15.43)/(2332.21/40 - 281.90/18)

=0.5086

Let level of significance be 5%

α=0.05tabulated at 5% level of significance is,

ztab=1.64

Zcal > Ztab

we reject H0 and conclude that, u1<u2

mean of population before is less than that of after.