A certain drug is used to treat asthma. In a clinical trial of the drug, 22 of 257 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? Two-tailed test Left-tailed test Right tailed test b. What is the test statistic? z= (Round to two decimal places as needed.) c. What is the P-value?

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A certain drug is used to treat asthma. In a clinical trial of the drug, 22 of 257 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below.

1-PropZTest
- prop < 0.1
- z = -0.769332196
- p = 0.220480664
- \(\hat{p} = 0.0856031128\)
- n = 257

a. Is the test two-tailed, left-tailed, or right-tailed?
- ○ Two-tailed test
- ○ Left-tailed test
- ○ Right-tailed test

b. What is the test statistic?
- z = [ ]
  (Round to two decimal places as needed.)

c. What is the P-value?
- P-value = [ ]
  (Round to four decimal places as needed.)

d. What is the null hypothesis, and what do you conclude about it?
- Identify the null hypothesis.
  - ○ A. \(H_0: p \neq 0.1\)

The calculator display provides the results for hypothesis testing regarding the proportion of treated subjects experiencing headaches, with the following aspects:

- **z-value**: The calculated z-value of -0.7693 indicates the position of the sample proportion in terms of standard deviations away from the hypothesized population proportion.
- **p-value**: Calculated as 0.2205, the p-value represents the probability of observing a sample statistic as extreme as the test statistic, under the null hypothesis.
  
This information assists in determining whether there's strong enough evidence to reject the null hypothesis that \( p = 0.1 \) at a significance level of 0.05.
Transcribed Image Text:A certain drug is used to treat asthma. In a clinical trial of the drug, 22 of 257 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. 1-PropZTest - prop < 0.1 - z = -0.769332196 - p = 0.220480664 - \(\hat{p} = 0.0856031128\) - n = 257 a. Is the test two-tailed, left-tailed, or right-tailed? - ○ Two-tailed test - ○ Left-tailed test - ○ Right-tailed test b. What is the test statistic? - z = [ ] (Round to two decimal places as needed.) c. What is the P-value? - P-value = [ ] (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? - Identify the null hypothesis. - ○ A. \(H_0: p \neq 0.1\) The calculator display provides the results for hypothesis testing regarding the proportion of treated subjects experiencing headaches, with the following aspects: - **z-value**: The calculated z-value of -0.7693 indicates the position of the sample proportion in terms of standard deviations away from the hypothesized population proportion. - **p-value**: Calculated as 0.2205, the p-value represents the probability of observing a sample statistic as extreme as the test statistic, under the null hypothesis. This information assists in determining whether there's strong enough evidence to reject the null hypothesis that \( p = 0.1 \) at a significance level of 0.05.
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