A politician claims that she will receive 65% of the vote in an upcoming election. The results of a properly designed random sample of 100 voters showed that 49 of those sampled will vote for her. Is it likely that her assertion is correct at the 0.05 level of significance?

Question

(a) State the appropriate null and alternative hypotheses.
Ho: p 0.65
Ha: p 0.65

(ii) Find z. (Give your answer correct to two decimal places.)


(iii) Find the p-value. (Give your answer correct to four decimal places.)

Answer 1

Let's try a binomial proportion one-sample z-test.

Ho: p = 0.65
Ha: p does not equal 0.65

Test statistic:
z = (0.49 - 0.65)/√[(0.65)(0.35)/100]
z = -3.35

The null would be rejected at the .05 level for a two-tailed test (p does not equal 0.65).

Use a z-table to find the p-value. The p-value is the actual level of the test statistic.