Spam: A researcher reported that $71.8 \%$ of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be $69 \%$. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that the percentage of emails that are spam is greater than $69 \%$ ? Use both $\alpha=0.01$ and $\alpha=0.05$ levels of significance and the $P$-value method with the TI- 84 Plus calculator.
Part: $0 / 5$
Part 1 of 5
(a) State the appropriate null and alternate hypotheses.
\[
\begin{array}{l}
H_{0}: p=0.69 \\
H_{1}: p< 0.69
\end{array}
\]

This hypotheses test is a left-tailed test.

Part: $1 / 5$

Part 2 of 5
(b) Compute the value of the test statistic. Round the answer to at least two decimal places.
$z=$
\[
1.93
\]

Part: $2 / 5$

Part 3 of 5
(c) Compute the $P$-value. Round the answer to at least four decimal places.
\[
P \text {-value }=\square
\]
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Step 1 ：Define the null and alternate hypotheses as follows: \(H_0: p = 0.69\) (The percentage of spam emails is 69%) and \(H_1: p > 0.69\) (The percentage of spam emails is greater than 69%).

Step 2 ：Identify that this is a right-tailed test.

Step 3 ：Calculate the sample proportion \(\hat{p}\) as \(\hat{p} = \frac{365}{500} = 0.73\).

Step 4 ：Calculate the test statistic z using the formula: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 * (1 - p_0)}{n}}}\).

Step 5 ：Substitute the values into the formula to get: \(z = \frac{0.73 - 0.69}{\sqrt{\frac{0.69 * (1 - 0.69)}{500}}} = 1.93\).

Step 6 ：The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. It can be found using a standard normal distribution table or a calculator with statistical functions.

Step 7 ：For a right-tailed test, the P-value is the area to the right of the test statistic.

Step 8 ：Given the z-score of 1.93, the P-value would be less than 0.05, indicating strong evidence against the null hypothesis at the \(\alpha = 0.05\) level of significance.

Step 9 ：For the \(\alpha = 0.01\) level of significance, we would need to compare the P-value to 0.01 to make a conclusion.

Step 10 ：Unfortunately, without a standard normal distribution table or a calculator with statistical functions, we can't calculate the exact P-value.