SOLUTION: The following hypotheses are given. H0 : π ≤ 0.83 H1 : π > 0.83 A sample of 140 observations revealed that p = 0.88. At the 0.05 significance level, can t

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Question 977800: The following hypotheses are given.
H0 : π ≤ 0.83
H1 : π > 0.83
A sample of 140 observations revealed that p = 0.88. At the 0.05 significance level, can the null hypothesis be rejected?
a. State the decision rule. (Round your answer to 2 decimal places.)
Reject H0 if z > ______?
b. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
Value of the test statistic=______?
c. What is your decision regarding the null hypothesis?
(reject/no not reject) H0.
Found 3 solutions by Boreal, stanbon, jim_thompson5910:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
standard error = sqrt{ (0.83)(0.17)/140}=0.03175
Reject Ho if z>= 1.645 (1 tailed test)
Test statistic is phat-p/SE= 0.05/0.03175
=1.57
Fail to reject Ho. There is insufficient evidence to conclude that the true proportion is >0.83.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The following hypotheses are given.
H0 : π ≤ 0.83
H1 : π > 0.83
A sample of 140 observations revealed that p-hat = 0.88.
At the 0.05 significance level, can the null hypothesis be rejected?
a. State the decision rule. (Round your answer to 2 decimal places.)
Reject H0 if z > invNorm(0.95 = 1.645?
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b. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
z(0.88) = (0.88-0.83)/sqrt[0.83*0.17/140] = 1.5750
Value of the test statistic = 1.5750
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c.What is your decision regarding the null hypothesis?
(reject/no not reject) H0.
Fail to reject Ho.
--------------
Cheers,
Stan H.
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Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
This is a right tailed test. So we want to find the value k that makes P(Z > k) = 0.05 true.
Use a table like this one. Look at the bottom Z row. Then find the column that has "0.05" in the "one tail row". The intersection of the row and column leads to 1.645

So 1.645 is the critical value we're after. That rounds to 1.65

Decision Rule: Reject H0 if z > 1.65

====================================================================
b)
Standard Error:
SE+=+sqrt%28%28pi%2A%281-pi%29%29%2Fn%29
SE+=+sqrt%28%280.83%2A%281-0.83%29%29%2F140%29
SE+=+0.03174676586452 (approximate)

Test Statistic
z+=+%28p+-+pi%29%2F%28SE%29
z+=+%280.88-0.83%29%2F%280.03174676586452%29
z+=+1.57496357938871
z+=+1.57

The test statistic is 1.57
====================================================================
c)

Go back to part a) where the decision rule is set up.

The critical value is 1.65 and the test statistic 1.57

Since the test statistic is NOT larger than the critical value, we fail to reject the null hypothesis.

We do not have enough evidence to reject the null.