Question 1077666: Help with p value, I entered tcdf(1.063,99,7), not working!
In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.
Are America's top chief executive officers (CEOs) really worth all that money? One way to answer this question is to look at row B, the annual company percentage increase in revenue, versus row A, the CEO's annual percentage salary increase in that same company. Suppose a random sample of companies yielded the following data:
B: Percent increase
for company 28 22 26 18 6 4 21 37
A: Percent increase
for CEO 15 19 28 14 -4 19 15 30
Do these data indicate that the population mean percentage increase in corporate revenue (row B) is different from the population mean percentage increase in CEO salary? Use a 5% level of significance.
(a) What is the level of significance?
.05
Correct: Your answer is correct.
State the null and alternate hypotheses.
H0: μd = 0; H1: μd > 0
H0: μd = 0; H1: μd < 0
H0: μd > 0; H1: μd = 0
H0: μd ≠ 0; H1: μd = 0
H0: μd = 0; H1: μd ≠ 0
Correct: Your answer is correct.
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that d has an approximately uniform distribution.
The Student's t. We assume that d has an approximately uniform distribution.
The Student's t. We assume that d has an approximately normal distribution.
The standard normal. We assume that d has an approximately normal distribution.
Correct: Your answer is correct.
What is the value of the sample test statistic? (Round your answer to three decimal places.)
1.063
Correct: Your answer is correct.
(c) Find the P-value. (Round your answer to four decimal places.)
.1615
Incorrect: Your answer is incorrect.
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! There are three possibilities.
When I use a calculator, I get a t-statistic of 0.61, which is not the 1.063 called correct. My p-value is 0.5536, which is quite different as well.
The other issue is that the df=14 in this test, which doesn't change the p-value a lot, but does change it to about 0.1529.
A third possibility, and the most important, is that a one way p-value has to be doubled for a two way test. This is a 2-way test as I interpret it, so the amount goes on both sides of the curve, and the p-value would be 0.3230. The calculator and the table give the right hand interval only using the numbers provided.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i used a paired sample t-test calculator to duplicate your results.
what i got was:
t = 1.062965
i then looked for the p value.
what the calculator told me was:
p = 0.161543 if i assumed a one tail test.
p = 0.323087 if i assumed a two tail test.
if you double .161543, you will get .323086 which i assume is equivalent to .323087 if you don't assume any truncating of decimal places.
because this is a two tailed test, what this is saying is that your distribution is symmetric and your p-value is your tail on both ends of the distribution, which is why the tail is equal to 2 times what it would be if it was a one tail test.
i'm not sure if this is accurate, but they do show it this way.
the calculator i used was at this link:
http://www.socscistatistics.com/tests/ttestdependent/Default2.aspx
the quckest way to see if this is what they are assuming is to enter the p-value as .323087 or .3231 rounded to 4 decimal places and see if they agree that is the correct answer.
in either case, the p-value is greater than the critical p-value which you should have calculated was a t-value of
i believe the theory is that .05 level of significance is two tailed level os significance, so you have to double the p-value of the t-score to compare.
the other side of the coin is to compare the p-value against .025 rather than .05. .1615 is still greater so the null hypothesis cannot be rejected.
i don't usually use the p-value.
i prefer using the t-value.
the critical t-value for 7 degrees of freedom at .05 alpha in a two sides test is 2.365.
the t-score of 1.06... is well below that, so the null hypothesis can't be rejected.
try doubling the p-value you calculated and see if that's accepted as correct.
when dealing with p-values, i always compare one side only, i.e. alpha / 2 = .025 which is the critical p-value on the high side.
the p-value of the t-score of 1.06... is .1615 on the high side.
can't reject the null hypotheses since it is greater than .025.
bottom line is you are either comparing .1615 to .025 or you are comparing .323... to .05.
at least that's what i think is happening.
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