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 70%. If 158 out of a random sample of 235 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 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. (If necessary, consult a list of formulas.)

The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF The value of the test statistic: (Round to at least three decimal places.) The critical value at the 0.1 level of significance: (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 70% ? Yes No

Answer 1

We have to test, H0: p 0.7 against H1: p < 0.7

The test-statistic is given by, Z = , where, = 158/235 = 0.6723, = 0.7, n = 235

Hence, Z = - 0.9266

Under H0, Z ~ N(0,1)

The p-value = P(Z < - 0.9266) = (-0.9266) = 0.1771

Since, p-value > level of significance, we fail to reject the null hypothesis H0.

We conclude there is not sufficient evidence to reject the claim that the proportion is less than 70%.

The null hypothesis	H0: p 0.7
The alternative hypothesis	H1: p > 0.7
The type of test-statistic	Z-statistic
The value of the test-statistic	-0.9266
The critical value at 0.1 level of significance	-1.282
Decision of Rejecting the aides' estimate	No, fail to reject H0