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?

(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.)

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.

(i) The appropriate null hypothesis (Ho) is that the politician's claim is true, so the proportion of voters who will vote for her is p = 0.65. The alternative hypothesis (Ha) is that her claim is not true, so the proportion of voters who will vote for her is p ≠ 0.65.

(ii) To calculate z, we need the sample proportion (p̂) and the standard error (SE). The sample proportion is the number of voters sampled who will vote for her divided by the sample size: p̂ = 49/100 = 0.49.

The standard error can be calculated using this formula: SE = √(p̂ * (1 - p̂) / n), where n is the sample size. In this case, SE = √(0.49 * (1 - 0.49) / 100) ≈ 0.0494.

Now we can find z by subtracting the hypothesized proportion (p = 0.65) from the sample proportion (p̂ = 0.49) and dividing by the standard error: z = (p̂ - p) / SE = (0.49 - 0.65) / 0.0494 ≈ -3.24.

(iii) The p-value represents the probability of obtaining a sample proportion as extreme or more extreme than the one observed, assuming the null hypothesis is true. In this case, we use a two-tailed test because the alternative hypothesis is p ≠ 0.65.

To find the p-value, we look up the absolute value of z (|z| = 3.24) in the standard normal distribution table. The p-value for a z-score of 3.24 is approximately 0.0012.

Therefore, the p-value is 0.0012 (correct to four decimal places).