Math Problem Statement

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 77%. He examines a random sample of 500 emails received at an email server, and finds that 361 of the messages are spam. Can you conclude that less than 77% of emails are spam? Use both =α0.05 and =α0.10 levels of significance and the critical value method with the table. State the appropriate null and alternate hypotheses. H0: H1: This hypothesis test is a (Left or right-tailed or two-tailed) test.

Solution

Step-by-step Hypothesis Testing Process:

We are tasked with determining if the proportion of spam emails in a company's sample is less than 77%. Here's how we can approach it:

1. State the Null and Alternative Hypotheses:

  • Null hypothesis (H₀): The proportion of spam emails is greater than or equal to 77% (0.77).

    H0:p0.77H_0: p \geq 0.77

  • Alternative hypothesis (H₁): The proportion of spam emails is less than 77% (0.77).

    H1:p<0.77H_1: p < 0.77

This is a left-tailed test because we are testing whether the proportion is less than 0.77.

2. Significance Levels (α):

We are asked to use two significance levels:

  • α=0.05\alpha = 0.05
  • α=0.10\alpha = 0.10

3. Sample Proportion (p̂):

We are given:

  • Number of spam emails in the sample = 361
  • Sample size n=500n = 500

The sample proportion p^\hat{p} is calculated as:

p^=361500=0.722\hat{p} = \frac{361}{500} = 0.722

4. Test Statistic:

The test statistic for a proportion follows a z-distribution and is calculated as:

z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}

Where:

  • p^=0.722\hat{p} = 0.722 (sample proportion)
  • p0=0.77p_0 = 0.77 (hypothesized population proportion)
  • n=500n = 500 (sample size)

Substituting the values:

z=0.7220.770.77(10.77)500=0.0480.77(0.23)500z = \frac{0.722 - 0.77}{\sqrt{\frac{0.77(1 - 0.77)}{500}}} = \frac{-0.048}{\sqrt{\frac{0.77(0.23)}{500}}}

z=0.0480.1771500=0.0480.0003542=0.0480.018822.55z = \frac{-0.048}{\sqrt{\frac{0.1771}{500}}} = \frac{-0.048}{\sqrt{0.0003542}} = \frac{-0.048}{0.01882} \approx -2.55

5. Critical Value Method:

Now we compare the test statistic z=2.55z = -2.55 to the critical values for each significance level.

  • For α=0.05\alpha = 0.05, the critical value for a left-tailed test is zα=1.645z_{\alpha} = -1.645.
  • For α=0.10\alpha = 0.10, the critical value is zα=1.28z_{\alpha} = -1.28.

6. Decision:

  • At α=0.05\alpha = 0.05, since z=2.55<1.645z = -2.55 < -1.645, we reject the null hypothesis.
  • At α=0.10\alpha = 0.10, since z=2.55<1.28z = -2.55 < -1.28, we reject the null hypothesis.

7. Conclusion:

There is sufficient evidence to conclude, at both the 5% and 10% significance levels, that the proportion of spam emails in this company is less than 77%.


Would you like a detailed explanation of any specific part? Here are 5 follow-up questions you could explore:

  1. How do we interpret the significance levels in hypothesis testing?
  2. What is the role of the sample size in determining the test statistic?
  3. How does the choice of one-tailed vs. two-tailed tests impact the critical value?
  4. What are Type I and Type II errors in the context of this problem?
  5. How would the conclusion change if we used a higher or lower significance level?

Tip: In hypothesis testing, always ensure that the test type (left, right, or two-tailed) matches the hypothesis you're testing to avoid incorrect conclusions.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Significance Levels
Critical Value Method

Formulas

Sample Proportion p̂ = x/n
Z-test for Proportion: z = (p̂ - p₀) / √(p₀(1 - p₀) / n)

Theorems

Z-distribution

Suitable Grade Level

Undergraduate