Question 1198699: 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 265 students at this college is selected, and it is found that 46 commute more than fifteen miles to school, can we reject the college's claim at the 0.01 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.
a) The null hypothesis:
b) The alternative hypothesis:
c) The type of test statistic (Z, t, chi square, or f):
d) The Value of the test statistic (round to at least 3 decimal places):
e) The critical value (round to at least 3 decimal places):
f) Is there enough evidence to support the claim that the proportion of students who commute more than fifteen miles to school is less than 25%?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population p = .25
sample p = 46 / 265 = .1735849057
.01 one tailed confidence interval = .01
standard error = sqrt(.25 * .75 / 265) = .0265997588.
z-score = (.1735849057 - .25) / .0265997588 = -2.872773952.
that's the area to the left of the z-score.
the probability of getting a z-score less than .17..... is equal to the area to the left of that z-score which is equal to .002.
since that is less than .01, the results are significant, supporting the claim that the proportion of students who commute more than 15 miles to school is no more than 25%.
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