In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6955 subjects randomly selected from an online group involved with ears. There were 1310 surveys retumed. 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. Identify the null hypothesis and alternative hypothesis. O A. Ho p>0.2 H p=0.2 O B. Ho p=0.2 H,: p>0.2 YO D. Ho p=0.2 H, p<0.2 OC. Ho p<0.2 H p=02 O E. Ho p=0.2 H, p#0.2 OF. Ho p#0.2 H p=0.2 The test statistic is z = (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) hat the return rate is less than 20%

MATLAB: An Introduction with Applications
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In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6,955 subjects randomly selected from an online group involved with ears. There were 1,310 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.

### Identify the null hypothesis and alternative hypothesis.

- **A.** \( H_0: p > 0.2 \)  
  \( H_1: p = 0.2 \)

- **B.** \( H_0: p = 0.2 \)  
  \( H_1: p > 0.2 \)

- **C.** \( H_0: p < 0.2 \)  
  \( H_1: p = 0.2 \)

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

- **E.** \( H_0: p = 0.2 \)  
  \( H_1: p \neq 0.2 \)

- **F.** \( H_0: p \neq 0.2 \)  
  \( H_1: p = 0.2 \)

The test statistic is \( z = \_\_\_ \)  
(Round to two decimal places as needed.)

The P-value is \( \_\_\_ \)  
(Round to three decimal places as needed.)

Because the P-value is \_\_\_ the significance level \_\_\_, \_\_\_ the null hypothesis. There is \_\_\_ evidence to support the claim that the return rate is less than 20%.
Transcribed Image Text:In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6,955 subjects randomly selected from an online group involved with ears. There were 1,310 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. ### Identify the null hypothesis and alternative hypothesis. - **A.** \( H_0: p > 0.2 \) \( H_1: p = 0.2 \) - **B.** \( H_0: p = 0.2 \) \( H_1: p > 0.2 \) - **C.** \( H_0: p < 0.2 \) \( H_1: p = 0.2 \) - **D.** \( H_0: p = 0.2 \) \( H_1: p < 0.2 \) - **E.** \( H_0: p = 0.2 \) \( H_1: p \neq 0.2 \) - **F.** \( H_0: p \neq 0.2 \) \( H_1: p = 0.2 \) The test statistic is \( z = \_\_\_ \) (Round to two decimal places as needed.) The P-value is \( \_\_\_ \) (Round to three decimal places as needed.) Because the P-value is \_\_\_ the significance level \_\_\_, \_\_\_ the null hypothesis. There is \_\_\_ evidence to support the claim that the return rate is less than 20%.
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