Questions: Based on a survey, 35% of likely voters would be willing to vote by internet instead of the in-person traditional method of voting. For each of the following, assume that 12 likely voters are randomly selected. Complete parts (a) through (c) below. a. What is the probability that exactly 9 of those selected would do internet voting? 0.00476 (Round to five decimal places as needed.) b. If 9 of the selected voters would do internet voting, is 9 significantly high? Why or why not? Select the correct choice below and fill in the answer box within your choice. (Round to five decimal places as needed.) A. Yes, because the probability of 9 or more is . which is not low. B. No, because the probability of 9 or more is which is C. Yes, because the probability of 9 or more is 1. which is low. D. No, because the probability of 9 or more is which is not low.

Solution Steps

To find the probability that exactly 9 out of 12 selected voters would choose internet voting, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 12 \) (number of trials),
• \( x = 9 \) (number of successes),
• \( p = 0.35 \) (probability of success),
• \( q = 1 - p = 0.65 \) (probability of failure).

Calculating this gives:

\[ P(X = 9) = 0.00476 \]

Thus, the probability of exactly 9 voters choosing internet voting is \( 0.00476 \).

Next, we perform a hypothesis test to determine if having 9 voters choosing internet voting is significantly high. We set up the null hypothesis \( H_0: p = 0.35 \) and the alternative hypothesis \( H_a: p > 0.35 \).

The test statistic \( Z \) is calculated as follows:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

where:

• \( \hat{p} = \frac{9}{12} = 0.75 \) (sample proportion),
• \( p_0 = 0.35 \) (hypothesized population proportion),
• \( n = 12 \) (sample size).

Substituting the values, we find:

\[ Z = 2.90509 \]

The corresponding p-value for this test statistic is:

\[ \text{P-value} = 0.00184 \]

To determine if 9 is significantly high, we compare the p-value to the significance level \( \alpha = 0.05 \). Since \( 0.00184 < 0.05 \), we reject the null hypothesis.

Thus, we conclude that having 9 voters choosing internet voting is significantly high.

Final Answer