Question

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 71%. he examines a random sample of 500 emails received at an email server, and finds that 380 of the messages are spam. can you conclude that the percentage of emails that are spam differs from 71%? use both \\(\alpha = 0.10\\) and \\(\alpha = 0.01\\) levels of significance and the \\(p\\)-value method and excel.\
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part: 0 / 5\
part 1 of 5\
(a) state the appropriate null and alternate hypotheses.\
\\(h_0: p = 0.71\\)\
\\(h_1: p \
eq 0.71\\)\
this hypothesis test is a two - tailed test.\
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part: 1 / 5\
part 2 of 5\
(b) compute the test statistic. do not round intermediate steps. round the answer to two decimal places.\
\\(z = 2.46\\)\
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part: 2 / 5\
part 3 of 5\
compute the \\(p\\)-value. round the answer to four decimal places.\
\\(p\\)-value = 0.0139\
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part: 3 / 5\
part 4 of 5\
determine whether to reject \\(h_0\\).\
since \\(p \\ select\\ 0.10\\), we select the null hypothesis \\(h_0\\) at \\(\alpha = 0.10\\).\
since \\(p \\ select\\ 0.01\\), we select the null hypothesis \\(h_0\\) at \\(\alpha = 0.01\\).

Explanation:

Step1: Analyze α=0.10 case

Check if \( P\text{-value} < \alpha \). Given \( P\text{-value} = 0.0139 \) and \( \alpha = 0.10 \). Since \( 0.0139 < 0.10 \), we reject \( H_0 \).

Step2: Analyze α=0.01 case

Check if \( P\text{-value} < \alpha \). Given \( P\text{-value} = 0.0139 \) and \( \alpha = 0.01 \). Since \( 0.0139 > 0.01 \), we fail to reject \( H_0 \).

Answer:

For \( \alpha = 0.10 \): reject the null hypothesis \( H_0 \)
For \( \alpha = 0.01 \): fail to reject the null hypothesis \( H_0 \)