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 onetailed 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.

Solution :
This is the left tailed test .
The null and alternative hypothesis is
H_{0} : p 0.60
H_{a} : p < 0.60
= x / n = 111 / 210 = 0.5286
Test statistic = z
=  P_{0} / [P_{0 *} (1  P_{0} ) / n]
= 0.5286  0.60 / [(0.60 * 0.40) / 210]
= 2.113
P(z < 2.113) = 0.0173
Pvalue = 0.0173
= 0.01
Pvalue >
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% .
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