In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6995 subjects randomly selected from an online group involved with ears. There were 1326 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution.
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Part 1
Identify the null hypothesis and alternative hypothesis.
A.
Upper H 0​: pless than0.2
Upper H 1​: pequals0.2
B.
Upper H 0​: pnot equals0.2
Upper H 1​: pequals0.2
C.
Upper H 0​: pequals0.2
Upper H 1​: pgreater than0.2
D.
Upper H 0​: pequals0.2
Upper H 1​: pless than0.2
Your answer is correct.E.
Upper H 0​: pgreater than0.2
Upper H 1​: pequals0.2
F.
Upper H 0​: pequals0.2
Upper H 1​: pnot equals0.2
Part 2
The test statistic is zequals
  
enter your response here.
​(Round to two decimal places as​ needed.)


### Part 1: Identifying the Hypotheses

You are testing the claim that the return rate is less than 20%, so the hypotheses are:

- **Null Hypothesis (\(H_0\))**: \( p = 0.2 \)
- **Alternative Hypothesis (\(H_1\))**: \( p < 0.2 \)

So, the correct choice is:

**D. \( H_0: p = 0.2 \) and \( H_1: p < 0.2 \)**

### Part 2: Test Statistic Calculation

To calculate the test statistic, we can use the formula for the z-score in the context of proportions:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \(\hat{p}\) is the sample proportion.
- \(p_0\) is the hypothesized proportion (in this case, 0.2).
- \(n\) is the sample size.

#### Given Data:
- \( n = 6995 \)
- \( x = 1326 \) (number of returned surveys)
- \( p_0 = 0.2 \)

First, calculate the sample proportion \(\hat{p}\):

\[
\hat{p} = \frac{1326}{6995} \approx 0.1895
\]

Now, calculate the z-score:

\[
z = \frac{0.1895 - 0.2}{\sqrt{\frac{0.2 \times 0.8}{6995}}}
\]

Let's compute this.The test statistic \( z \) is approximately \(-2.18\).

### Summary
- **Null Hypothesis (\(H_0\))**: \( p = 0.2 \)
- **Alternative Hypothesis (\(H_1\))**: \( p < 0.2 \)
- **Test Statistic**: \( z \approx -2.18 \)

Would you like to calculate the \( p \)-value next or have any questions? 

Here are some related questions you might explore:
1. How do you interpret the p-value in this context?
2. What is the critical value for this test at the 0.01 significance level?
3. How does the choice of significance level affect your conclusion?
4. What assumptions must be met to use the normal distribution as an approximation in this test?
5. How would the results differ if the alternative hypothesis were \( p \neq 0.2 \)?

**Tip**: Always check the conditions before using the normal approximation for a binomial distribution!

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6995 subjects randomly selected from an online group involved with ears. There were 1326 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.

Test the claim that a survey's return rate is less than 20% using a 0.01 significance level and normal approximation. The problem explores hypothesis testing for proportions, with detailed steps to calculate the z-score and interpret results.

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