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 most 75 % . If 217 out of a random sample of 270 college students expressed an intent to vote, can the aide's estimate be rejected at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below.

Null Hypothesis

Alternative Hypothesis

type of test statistic

Value of the test statistic

the critical value at the .05 level of significance

Can we reject the aide's estimate that the proportion of college students who intend to vote is at most 75%?

Answer 1

answer)

Null hypothesis Ho : P <= 0.75

Alternate hypothesis Ha : P > 0.75

N = 270

P = 0.75

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 202.5

N*(1-p) = 67.5

Both the conditions are met so we can use standard normal z table to estimate the test.

Type of test statistics is z test statistics

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 217/270

Claimed P = 0.75

N = 270

Test statistics z = 2.038 ~ 2.04

Given alpha = 0.05

From z table,P(z>1.645)= 0.05

So rejection region is if test statistics is greater than 1.645 reject Ho

As the obtained test statistics is greater than 1.645

We reject the null hypothesis that p <=0.75

So, we do not have enough evidence to support the claim that the proportion of college students who intend to vote is at most 75%