Question

A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60%. If 111 out of a random sample of 210 college students expressed an intent to vote, can we reject the aide's estimate 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.

The null hypothesis:	H0:
The alternative hypothesis:	H1:
The type of test statistic:	(Choose one)Z, t, Chi square, F
The value of the test statistic: (Round to at least three decimal places.)
The p-value: (Round to at least three decimal places.)
Can we reject the aide's estimate that the proportion of college students who intend to vote is at least 60% ?		Yes	No

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p 0.60

H_a : p < 0.60

= x / n = 111 / 210 = 0.5286

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.5286 - 0.60 / [(0.60 * 0.40) / 210]

= -2.113

P(z < -2.113) = 0.0173

P-value = 0.0173

= 0.01

P-value >

Fail to reject the null hypothesis .

NO

We can not reject the aide's estimate that the proportion of college students who

intend to vote is at least 60% .