We appreciate your visit to A certain medication may cause side effects To further investigate this finding a researcher selects a separate random sample of 150 adults of which 32. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Sure, let's address this question step by step to determine if there is convincing statistical evidence that the true proportion of adults who experience side effects from the medication is greater than 0.15 at a significance level of [tex]\(\alpha = 0.05\)[/tex].
### Step 1: Define Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The true proportion of adults who experience side effects is [tex]\(p = 0.15\)[/tex].
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The true proportion of adults who experience side effects is [tex]\(p > 0.15\)[/tex].
### Step 2: Check Conditions for the Test
1. Random Condition: A random sample of 150 adults is selected, so this condition is met.
2. 10% Condition: We assume the sample of 150 is less than 10% of all adults who could take this medication, ensuring independence.
3. Large Counts Condition: Check if [tex]\(np_0 \geq 10\)[/tex] and [tex]\(n(1-p_0) \geq 10\)[/tex].
- [tex]\(np_0 = 150 \times 0.15 = 22.5\)[/tex]
- [tex]\(n(1-p_0) = 150 \times 0.85 = 127.5\)[/tex]
- Both are greater than 10, so this condition is met.
### Step 3: Calculate Sample Statistics
- Sample Size ([tex]\(n\)[/tex]): 150
- Number of Adults with Side Effects: 32
- Sample Proportion ([tex]\(\hat{p}\)[/tex]): [tex]\(\frac{32}{150} \approx 0.2133\)[/tex]
### Step 4: Calculate Standard Error and Z-score
- Standard Error (SE): [tex]\(\sqrt{\frac{0.15 \times (1-0.15)}{150}} \approx 0.0292\)[/tex]
- Z-score: [tex]\(\frac{\hat{p} - p_0}{SE} = \frac{0.2133 - 0.15}{0.0292} \approx 2.172\)[/tex]
### Step 5: Make a Conclusion
A Z-score of approximately 2.172 suggests that the sample proportion is 2.172 standard deviations above the hypothesized proportion. To find the statistical significance, we compare this Z-score to critical values from the standard normal distribution.
- Significance Level ([tex]\(\alpha\)[/tex]): 0.05
- A Z-score greater than approximately 1.645 (for one-tailed test at [tex]\(\alpha = 0.05\)[/tex]) provides evidence against the null hypothesis.
Since our calculated Z-score (2.172) is greater than 1.645, we reject the null hypothesis.
### Conclusion
There is convincing statistical evidence at the [tex]\(\alpha = 0.05\)[/tex] level to support that the true proportion of adults experiencing side effects from the medication is greater than 0.15.
### Step 1: Define Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The true proportion of adults who experience side effects is [tex]\(p = 0.15\)[/tex].
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The true proportion of adults who experience side effects is [tex]\(p > 0.15\)[/tex].
### Step 2: Check Conditions for the Test
1. Random Condition: A random sample of 150 adults is selected, so this condition is met.
2. 10% Condition: We assume the sample of 150 is less than 10% of all adults who could take this medication, ensuring independence.
3. Large Counts Condition: Check if [tex]\(np_0 \geq 10\)[/tex] and [tex]\(n(1-p_0) \geq 10\)[/tex].
- [tex]\(np_0 = 150 \times 0.15 = 22.5\)[/tex]
- [tex]\(n(1-p_0) = 150 \times 0.85 = 127.5\)[/tex]
- Both are greater than 10, so this condition is met.
### Step 3: Calculate Sample Statistics
- Sample Size ([tex]\(n\)[/tex]): 150
- Number of Adults with Side Effects: 32
- Sample Proportion ([tex]\(\hat{p}\)[/tex]): [tex]\(\frac{32}{150} \approx 0.2133\)[/tex]
### Step 4: Calculate Standard Error and Z-score
- Standard Error (SE): [tex]\(\sqrt{\frac{0.15 \times (1-0.15)}{150}} \approx 0.0292\)[/tex]
- Z-score: [tex]\(\frac{\hat{p} - p_0}{SE} = \frac{0.2133 - 0.15}{0.0292} \approx 2.172\)[/tex]
### Step 5: Make a Conclusion
A Z-score of approximately 2.172 suggests that the sample proportion is 2.172 standard deviations above the hypothesized proportion. To find the statistical significance, we compare this Z-score to critical values from the standard normal distribution.
- Significance Level ([tex]\(\alpha\)[/tex]): 0.05
- A Z-score greater than approximately 1.645 (for one-tailed test at [tex]\(\alpha = 0.05\)[/tex]) provides evidence against the null hypothesis.
Since our calculated Z-score (2.172) is greater than 1.645, we reject the null hypothesis.
### Conclusion
There is convincing statistical evidence at the [tex]\(\alpha = 0.05\)[/tex] level to support that the true proportion of adults experiencing side effects from the medication is greater than 0.15.
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