691 ***please show how to solve each problem using excel formulas*** Weekly Pay
ID: 3364039 • Letter: 6
Question
691
***please show how to solve each problem using excel formulas***
Weekly Pay 582 333 759 633 629 523 320 685 599 753 553 641 290 800 696 627 679 667 542 619 950 614 548 570 678 697 750 569 679 598 596 557 657 617 1230 648 760 804 675 736 565 587 565 687 498 712 533 424 772691
***please show how to solve each problem using excel formulas***
A recent issue of the AARP Bulletin reported that the average weekly pay for a woman with a high school degree was $520 (AARP Bulletin, January-February, 2010). Suppose you would like to determine if the average weekly pay for all working women is significantly greater than that for women with a high school degree. Data providing the weekly pay for a sample of 50 working women are available in the file named WeeklyPay. These data are consistent with the findings reported in the article mentioned above. State the hypotheses that should be used to test whether the mean weekly pay for all women is significantly greater than the mean weekly pay for women with a high a. school degree. b. Use the data in the file named WeeklyPay to compute the sample mean, the test sta- tistic, and the p-value. Use .05. Wh Repeat the hypothesis test using the critical value approach. at is your conclusion? d.Explanation / Answer
SolutionA:
null hypothesis:
Ho:=520
Alternative Hypothesis:
Ha:>520
alpha=0.05
Solutionb:
copy paste the data in Excel
in Cell 52 type as below to get sample mean
=AVERAGE(A2:A51)
sample mean=637.94
In cell 53 type as elow to get sample standard deviation
=STDEV.S(A2:A51)
=148.4694
n=50
We calculate t statistic as
t=samplemean-pop mean/sample sd/sqrt(n)
=637.94-520/148.4694/sqrt(50)
=5.62
t stat=5.62
degrees of freedom=n-1=50-1=49
to get p value in Excel type
=T.DIST.RT(5.62,49)
p=
4.47999E-07
p<0.05
Reject Null hypothesis.
There is sufficient evidence at 5% level of signiifcance to support the claim.
Solutionc:
p=0.0000
alpha=0.05
Decsion rule:
if p<0.05 reejct null hypothessis
if p>0.05 accept null hypothesis.
here p=0.0000
p<0.05
Reject Null hypothesis
4.47999E-07
p<0.05
Reject Null hypothesis.
There is sufficient evidence at 5% level of signiifcance to support the claim.
Solutiond:
df=49
alpha=0.05
alpha/2=0.05/2=0.025
t crit=2.009
t cal=5.62
t cal> t crit
Reject Null hypothesis
4.47999E-07
p<0.05
Reject Null hypothesis.
There is sufficient evidence at 5% level of signiifcance to support the claim.
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