Math Problem Statement

In​ 1997, a survey of 980 households showed that 164 of them use​ e-mail. Use those sample results to test the claim that more than​ 15% of households use​ e-mail. Use a 0.05 significance level. Use this information to answer the following questions. Question content area bottom Part 1 a. Which of the following is the hypothesis test to be​ conducted? A.Upper H 0 : p not equals 0.15 Upper H 1 : p equals 0.15 Upper H 0 : p not equals 0.15 Upper H 1 : p equals 0.15 B.Upper H 0 : p greater than 0.15 Upper H 1 : p equals 0.15 Upper H 0 : p greater than 0.15 Upper H 1 : p equals 0.15 C.Upper H 0 : p less than 0.15 Upper H 1 : p equals 0.15 Upper H 0 : p less than 0.15 Upper H 1 : p equals 0.15 D.Upper H 0 : p equals 0.15 Upper H 1 : p greater than 0.15 Upper H 0 : p equals 0.15 Upper H 1 : p greater than 0.15 E.Upper H 0 : p equals 0.15 Upper H 1 : p less than 0.15 Upper H 0 : p equals 0.15 Upper H 1 : p less than 0.15 F.Upper H 0 : p equals 0.15 Upper H 1 : p not equals 0.15 Upper H 0 : p equals 0.15 Upper H 1 : p not equals 0.15 Part 2 b. What is the test​ statistic? zequals

enter your response here ​(Round to two decimal places as​ needed.) Part 3 c. What is the​ P-value? ​P-valueequals

enter your response here ​(Round to three decimal places as​ needed.) Part 4 d. What is the​ conclusion? There is not sufficient evidence to support the claim that more than​ 15% of households use​ e-mail. There is sufficient evidence to support the claim that more than​ 15% of households use​ e-mail. Part 5 e. Is the conclusion valid​ today? Why or why​ not? A. ​No, the conclusion is not valid today because the population characteristics of the use of​ e-mail are changing rapidly. B. ​Yes, the conclusion is valid today because the requirements to perform the test are satisfied. C. You can make no decisions about the validity of the conclusion today. Which of the following is the hypothesis test to be conducted?z(Round to two decimal places as needed.)P-value(Round to three decimal places as needed.)What is the conclusion?Why or why not?

Solution

To test the claim that more than 15% of households use e-mail, we will perform a hypothesis test for a population proportion pp. Here's how to answer each part of the problem:


Part 1: Hypotheses

We are testing the claim that more than 15% of households use e-mail. This leads to the following hypotheses:

  • Null Hypothesis (H0H_0): p=0.15p = 0.15 (The proportion of households that use e-mail is 15%.)
  • Alternative Hypothesis (H1H_1): p>0.15p > 0.15 (The proportion of households that use e-mail is more than 15%.)

Thus, the correct option is:

D. H0:p=0.15H_0: p = 0.15 and H1:p>0.15H_1: p > 0.15


Part 2: Test Statistic

We can calculate the test statistic using the formula for the z-score in a hypothesis test for a proportion:

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

Where:

  • p^\hat{p} is the sample proportion,
  • p0p_0 is the hypothesized population proportion (0.15),
  • nn is the sample size (980).
  1. Sample proportion p^\hat{p}: p^=164980=0.1673\hat{p} = \frac{164}{980} = 0.1673

  2. Now calculate the z-statistic: z=0.16730.150.15(10.15)980z = \frac{0.1673 - 0.15}{\sqrt{\frac{0.15(1 - 0.15)}{980}}}

Let me calculate this for you:

z=0.16730.150.15(0.85)980z = \frac{0.1673 - 0.15}{\sqrt{\frac{0.15(0.85)}{980}}}

Calculating zz-value:

z1.05z \approx 1.05

Thus, the test statistic z1.05z \approx 1.05.


Part 3: P-value

The P-value corresponds to the probability of observing a test statistic as extreme as the one computed under the null hypothesis. Since this is a one-tailed test, we will find the area to the right of the z-value from the standard normal distribution.

Using a standard normal table or calculator for z=1.05z = 1.05:

The P-value 0.147\approx 0.147.


Part 4: Conclusion

To draw a conclusion, we compare the P-value to the significance level α=0.05\alpha = 0.05.

  • If PP-value <α< \alpha, we reject the null hypothesis.
  • If PP-value α\geq \alpha, we fail to reject the null hypothesis.

Since P=0.147P = 0.147 is greater than 0.050.05, we fail to reject the null hypothesis.

Conclusion: There is not sufficient evidence to support the claim that more than 15% of households use e-mail.


Part 5: Is the conclusion valid today?

Given that the data is from 1997 and the characteristics of e-mail usage have changed significantly since then, the conclusion is likely not valid today.

So, the correct option is:

A. No, the conclusion is not valid today because the population characteristics of the use of e-mail are changing rapidly.


Summary:

  1. Hypotheses: H0:p=0.15H_0: p = 0.15, H1:p>0.15H_1: p > 0.15.
  2. Test statistic: z1.05z \approx 1.05.
  3. P-value: 0.147\approx 0.147.
  4. Conclusion: There is not sufficient evidence to support the claim that more than 15% of households use e-mail.
  5. Validity today: No, the conclusion is not valid today due to changing population characteristics.

Do you want further clarification or any additional details?

Here are 5 related questions:

  1. What conditions must be met to perform a hypothesis test for a population proportion?
  2. How is the critical value method different from the P-value method?
  3. What changes would you expect if you conducted the same test with modern data?
  4. How would the test change if the sample size were larger or smaller?
  5. What is the relationship between confidence intervals and hypothesis testing?

Tip: Always verify if the assumptions of normality and randomness in sampling are satisfied before conducting a hypothesis test for proportions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Statistical Inference

Formulas

z = (p̂ - p0) / √[p0(1 - p0) / n]

Theorems

Central Limit Theorem
Normal Approximation for Proportions

Suitable Grade Level

Grades 11-12 or College Level (Statistics)