SOLUTION: A recent study at a local college claimed that the proportion, p , of students who commute more than fifteen miles to school is no more than 25% . If a random sample of 255 student

Question 457582: A recent study at a local college claimed that the proportion, p , of students who commute more than fifteen miles to school is no more than 25% . If a random sample of 255 students at this college is selected, and it is found that 67 commute more than fifteen miles to school, can we reject the college's claim at the 0.05 level of significance?
Perform a one-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.
g) The null hypothesis:
h) The alternative hypothesis:
i) The type of test statistic (Z, t, chi square, or f):
j) The Value of the test statistic (round to at least 3 decimal places):
k) The p-value (round to at least 3 decimal places):

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A recent study at a local college claimed that the proportion, p , of students who commute more than fifteen miles to school is no more than 25% .
If a random sample of 255 students at this college is selected, and it is found that 67 commute more than fifteen miles to school, can we reject the college's claim at the 0.05 level of significance?
----------------
Perform a one-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.
g) The null hypothesis:
Ho: p <= 0.25 (claim)
h) The alternative hypothesis:
Ha: p > 0.25
---------------------
i) The type of test statistic (Z, t, chi square, or f):
Ans: z
-------------
j) The Value of the test statistic (round to at least 3 decimal places):
z(67/225) = (0.2978 - 0.25)/sqrt[0.25*0.75] = 0.1103
----------------------
k) The p-value (round to at least 3 decimal places):
p-value = P(z > 0.1103) = 0.4561
Conclusion: Since the p-value is greater than 5%,
fail to reject Ho.
The test results support the claim.
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Cheers,
Stan H.
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