You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 0.58. You use a significance level of [tex]\alpha=0.005[/tex].

[tex]\[
\begin{array}{l}
H_0: p=0.58 \\
H_1: p\ \textgreater \ 0.58
\end{array}
\][/tex]

You obtain a sample of size [tex]n=434[/tex] in which there are 266 successes.

1. What is the test statistic for this sample? (Report answer accurate to three decimal places.)

Test statistic [tex]= \square[/tex]

2. What is the [tex]p[/tex]-value for this sample? (Report answer accurate to four decimal places.)

[tex]p[/tex]-value [tex]= \square[/tex]



Answer :

Let's go through the problem step-by-step to determine the test statistic and the p-value for this sample.

1. State the hypotheses:
- Null hypothesis (\(H_0\)): The proportion of voters who prefer Candidate A is \( p_0 = 0.58 \).
- Alternative hypothesis (\(H_1\)): The proportion of voters who prefer Candidate A is greater than 0.58 (\( p > 0.58 \)).

2. Given data:
- Sample size (\(n\)): 434
- Number of successes (\(k\)): 266
- Significance level (\(\alpha\)): 0.005
- Population proportion under the null hypothesis (\(p_0\)): 0.58

3. Sample proportion (\(\hat{p}\)):
[tex]\[ \hat{p} = \frac{k}{n} = \frac{266}{434} = 0.6138 \][/tex]

4. Calculate the standard error (SE) of the sample proportion:
[tex]\[ \text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.58 \cdot (1 - 0.58)}{434}} = \sqrt{\frac{0.58 \cdot 0.42}{434}} = \sqrt{\frac{0.2436}{434}} = \sqrt{0.000561} \approx 0.0237 \][/tex]

5. Calculate the test statistic (z):
[tex]\[ z = \frac{\hat{p} - p_0}{\text{SE}} = \frac{0.6138 - 0.58}{0.0237} \approx \frac{0.0338}{0.0237} \approx 1.389 \][/tex]

So, the test statistic \(z\) is approximately \(1.389\).

6. Determine the p-value:
The p-value corresponds to the probability of getting a value as extreme as, or more extreme than, the observed value of the test statistic under the null hypothesis.

Since we are dealing with a right-tailed test:
[tex]\[ p\text{-value} = 1 - \Phi(z) \approx 1 - \Phi(1.389) \][/tex]

Looking up \( \Phi(1.389) \) in the standard normal distribution table (or using a calculator/software):
[tex]\[ \Phi(1.389) \approx 0.9176 \][/tex]

Hence:
[tex]\[ p\text{-value} = 1 - 0.9176 = 0.0824 \][/tex]

So, the results are:

- Test statistic \( \approx 1.389 \)
- \( p \)-value \( \approx 0.0824 \)

Therefore, the test statistic is [tex]\(1.389\)[/tex] (accurate to three decimal places) and the [tex]\( p \)[/tex]-value for this sample is [tex]\(0.0824\)[/tex] (accurate to four decimal places).