Question 1191810: You wish to test the following claim (Ha) at a significance level of α=0.005
Ho: p1 = p2
Ha: p1 < p2
You obtain 86.4% successes in a sample of size n1=766 from the first population. You obtain 91.1% successes in a sample of size n2=649 from the second population.
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =
The p-value is...
-less than (or equal to)α
-greater than α
This test statistic leads to a decision to...
-reject the null
-accept the null
-fail to reject the null
As such, the final conclusion is that...
(a)There is sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.
(b)There is not sufficient evidence to warrant rejection of the claim that the first population proportion is less than the second population proportion.
(c)The sample data support the claim that the first population proportion is less than the second population proportion.
(d)There is not sufficient sample evidence to support the claim that the first population proportion is less than the second population proportion.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
We're going to use a Two Proportion Z-Test.
Hypotheses:
Ho: p1 = p2
Ha: p1 < p2
The claim is in the alternative hypothesis and the claim is p1 < p2
The inequality sign in the alternative hypothesis says we have a left-tailed test.
Sample sizes
n1 = 766
n2 = 649
Sample proportions
phat1 = 0.864
phat2 = 0.911
phat = x/n
x = n*phat
x1 = n1*phat1
x1 = 766*0.864
x1 = 661.824
x1 = 662
x2 = n2*phat2
x2 = 649*0.911
x2 = 591.239
x2 = 591
pbar = pooled or average sample proportion
pbar = (x1+x2)/(n1+n2)
pbar = (662+591)/(766+649)
pbar = 0.885512
Think of pbar in the same light as something like xbar
It's the letter p with a horizontal bar over top.
SE = standard error
SE = sqrt( pbar*(1-pbar)*(1/n1 + 1/n2) )
SE = sqrt( 0.885512*(1-0.885512)*(1/766 + 1/649) )
SE = 0.016987
z = test statistic
z = (phat1 - phat2)/SE
z = (0.864 - 0.911)/0.016987
z = -2.767
Since the test statistic is recorded to three decimal places, we can't use a standard Z table unfortunately.
Instead, we can only use a p-value calculator.
There are countless many such free calculators out there, so feel free to pick your favorite.
One such option is this really neat online calculator
https://davidmlane.com/normal.html
Make sure the mean and standard deviation are 0 and 1 respectively.
The p-value you should get is roughly 0.0028
Keep in mind this is a left-tailed test.
Comparing this to the level of significance alpha = 0.005, we see that the p-value is smaller.
Whenever the p-value is smaller than alpha, we reject the null.
A useful saying is "If the p-value is low, then the null must go".
Since we rejected the null, we accept the claim that p1 < p2.
Translating back to the original problem, we have the conclusion that
(c)The sample data support the claim that the first population proportion is less than the second population proportion.
Once again, the claim is in the alternative hypothesis.
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Answers:
test statistic = -2.767
p-value = 0.0028
The p-value is less than alpha = 0.005
It means we reject the null
Conclusion:
(c)The sample data support the claim that the first population proportion is less than the second population proportion.
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