litician claims that she will receive \( 60 \% \) of the votes in the upcoming election, of a rand ple of 200 voters, there were 100 who will surely vote for her. Test the politician's asser 05 level of significance.

To test the politician's assertion, we can conduct a hypothesis test. The null hypothesis \( H_0 \) is that she will indeed receive \( 60 \% \) of the votes, which corresponds to an expected count of \( 0.60 \times 200 = 120 \) votes. The alternative hypothesis \( H_a \) posits that she will receive less than \( 60 \% \), leading to \( 120 \) votes being the benchmark. Given that 100 voters are already determined to vote for her, we can use a one-sample z-test for proportions. The test statistic can be calculated using the formula: \[ z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Here, \( \hat{p} \) is the sample proportion (i.e., \( \frac{100}{200} = 0.5 \)), \( p_0 = 0.6 \), and \( n = 200 \). Plugging in the values, we get: \[ z = \frac{(0.5 - 0.6)}{\sqrt{\frac{0.6(0.4)}{200}}} \approx \frac{-0.1}{0.03464} \approx -2.89 \] Next, we can look up this z-value in the standard normal distribution table. At a \( 0.05 \) level of significance for a one-tailed test, the critical z-value is approximately \( -1.645 \). Since \( -2.89 < -1.645 \), we reject the null hypothesis. The evidence suggests that the politician may not achieve her claimed \( 60 \% \) of votes based on the given sample data. Significant conclusion: We conclude that the politician's claim may be overly optimistic based on the sample of voters surveyed!