Final answer:
Calculating the sample proportion, standard error, and z-score, we find a p-value of approximately 0.0806. Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
Step-by-step explanation:
To test the candidate's claim and determine if less than 5.8% of the ballots were defective, we can use a hypothesis test. Let's assume the null hypothesis (H-o) that less than or equal to 5.8% of the ballots were defective and the alternative hypothesis (Ha) that more than 5.8% were defective.
Using the information given, we have a sample of 500 ballots, out of which 21 are defective. We can calculate the sample proportion of defective ballots by dividing the number of defective ballots by the total sample size.
Sample Proportion (p) = Number of defective ballots / Sample size
p = 21/500 = 0.042
Next, we can find the standard error of the sample proportion using the formula:
Standard Error (SE) = sqrt((p*(1-p)) / n)
SE = sqrt((0.042*(1-0.042)) / 500) = 0.00917
To test the claim, we can calculate the z-score using the formula:
Z-score = (Sample Proportion - Hypothesized Proportion) / Standard Error
= (0.042 - 0.058) / 0.00917 = -1.746
Finally, we can find the p-value associated with the z-score using a standard normal distribution table or using Excel.
The p-value for a z-score of -1.746 is approximately 0.0806 (rounded to three decimal places).
Since the p-value (0.0806) is greater than the significance level (0.05), we fail to reject the null hypothesis. Therefore, there is not enough evidence to support the candidate's claim that more than 5.8% of the ballots were defective.