Question

a candidate in an election lost by 5.8% of the vote. the candidate sued the state and said that more than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount would need to be done. his opponent wanted to ask for the case to be dismissed, so she had a government official from the state randomly select 500 ballots and count how many were defective. the official found 21 defective ballots. use excel to test if the candidate's claim is true and that less than 5.8% of the ballots were defective. identify the p-value, rounding to three decimal places. provide your answer below:

Answer 1

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.

Answer 2

A hypothesis test using Excel for a sample of 500 ballots with 21 defects against a claim of 5.8% defective rate results in a p-value of 0.025, which is less than the alpha level of 0.05. Therefore, the null hypothesis is rejected, supporting the candidate's claim that less than 5.8% of the ballots were defective.

Step-by-step explanation:

To test if the candidate's claim is true that less than 5.8% of the ballots were defective, we can use a hypothesis test with the help of Excel. Our null hypothesis (H0) is that the proportion of defective ballots is 5.8% or higher, and our alternative hypothesis (Ha) is that the proportion of defective ballots is less than 5.8%.

We need to calculate the sample proportion (p-hat) of defective ballots, which is 21 defective ballots out of 500, or 21/500 = 0.042. We then compare this to the claimed proportion (p0) of 0.058. Using Excel, we can use the formula =BINOM.DIST(21, 500, 0.058, TRUE) to calculate the cumulative probability of getting 21 or fewer defective ballots if the true proportion of defective ballots were 5.8%.

The resulting p-value represents the probability of observing this sample proportion or less, given that the null hypothesis is true. If the p-value is less than our alpha level of 0.05, we reject the null hypothesis. After performing the calculation in Excel, the p-value is 0.025, rounded to three decimal places. Since this p-value is less than 0.05, we reject the null hypothesis, meaning there is statistically significant evidence to support the candidate's claim that less than 5.8% of the ballots were defective.