Question 1202466: A college claims that the proportion, p, of students who commute more than fifteen miles to school is less than 25%. A researcher wants to test this. A random sample of 265 students at this college is selected, and it is found that 58 commute more than fifteen miles to school. Is there enough evidence to support the college's claim at the 0.05 level of significance?
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis H_{0} and the alternative hypothesis H_{1}
(b) Determine the type of test statistic to use. (Z, t, F, or chi square)
(c) Find the value of the test statistic. (Round to three or more decimal places.)
(d) Find the p-value. (Round to three or more decimal places.)
(e) 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%?
Yes or No
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! claim is that less than 25% commute more than 15 miles to school.
sample size is 265.
in that sample, 58 commute more than 15 miles.
alpha is .05 on one end of the confidence interval.
the confidence interval is equal to .95.
null hypothesis is tht the proportion of students who commute more than fiften miles to school is less than .25.
alternate hypothesis is that the proportion of student who commute more than fifteen miles to school is not less than .25.
the alpha will be on the high end of the confidence interval because the test is looking for a higher proportion than the assumed population proportion (i believe).
population p is assumed to be less than .25.
the mean proportion of the test would be .25.
sample p is equal to 58 / 265 = .2188679245.
population q is equal to 1 - .25 = .75
sample q is e4qual to 1 - .2188679245 = .7811320755
standard error of the test is square root of (population p * population q / 265) = .0265997588.
critical z-score at alpha = .05 would be z = 1.64485 rounded to 5 decimal places.
test z-score formula used is z = (x - m) / s
z is the z-score
x is the sample proportion.
m is the assumed mean proportion.
s is the standard error.
formula becomes z = (.2188679245 - .25) / .0265997588 = -1.170389389.
area to the right of that z-score is equal to .8790777911.
that's your test alpha on the high end.
it's not even close.
there is clearly not enough evidence to support the alternate hypothesis claim that the proportion of studdents who commute more than fifteen miles to school is not less than .25.
|
|
|
