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Answer :
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### State
We want to determine if there is convincing statistical evidence that the true proportion of adults who would experience side effects from this medication is greater than 0.15.
Hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\( p = 0.15 \)[/tex] (the true proportion of adults experiencing side effects is 0.15)
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\( p > 0.15 \)[/tex] (the true proportion is greater than 0.15)
Significance Level:
- [tex]\(\alpha = 0.05\)[/tex]
### Plan
We will use a one-sample z-test for the proportion to test these hypotheses. Let's check if the necessary conditions are satisfied:
1. Random Condition:
- The sample of 150 adults was selected randomly, so this condition is met.
2. 10% Condition:
- This condition states that the sample size should be no more than 10% of the population of interest. If the sample (150 adults) is less than 10% of all adults taking the medication, then this condition is satisfied. Typically, we assume this is met unless further information suggests otherwise.
3. Large Counts Condition:
- To ensure the sampling distribution of the sample proportion is approximately normal, we check that [tex]\( n \times p_0 \times (1-p_0) \geq 10 \)[/tex].
- Here, [tex]\( n = 150 \)[/tex] and [tex]\( p_0 = 0.15 \)[/tex].
Calculating:
[tex]\[
n \times p_0 \times (1-p_0) = 150 \times 0.15 \times 0.85 = 19.125
\][/tex]
Since 19.125 is greater than 10, the large counts condition is met.
With these conditions verified, we can proceed with the z-test for one proportion.
### Verification of Statements
- [tex]\(H_0: p=0.15\)[/tex]: True
- [tex]\(H_{a}: p<0.15\)[/tex]: False (the correct alternative hypothesis [tex]\(H_a\)[/tex] is [tex]\(p > 0.15\)[/tex])
- The random condition is met: True
- The 10% condition is met: True
- The large counts condition is met: True
- The test is a [tex]\(z\)[/tex]-test for one proportion: True
Thus, the true statements based on the above analysis are:
- [tex]\(H_0: p=0.15\)[/tex]
- The random condition is met.
- The 10% condition is met.
- The large counts condition is met.
- The test is a [tex]\(z\)[/tex]-test for one proportion.
### State
We want to determine if there is convincing statistical evidence that the true proportion of adults who would experience side effects from this medication is greater than 0.15.
Hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\( p = 0.15 \)[/tex] (the true proportion of adults experiencing side effects is 0.15)
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\( p > 0.15 \)[/tex] (the true proportion is greater than 0.15)
Significance Level:
- [tex]\(\alpha = 0.05\)[/tex]
### Plan
We will use a one-sample z-test for the proportion to test these hypotheses. Let's check if the necessary conditions are satisfied:
1. Random Condition:
- The sample of 150 adults was selected randomly, so this condition is met.
2. 10% Condition:
- This condition states that the sample size should be no more than 10% of the population of interest. If the sample (150 adults) is less than 10% of all adults taking the medication, then this condition is satisfied. Typically, we assume this is met unless further information suggests otherwise.
3. Large Counts Condition:
- To ensure the sampling distribution of the sample proportion is approximately normal, we check that [tex]\( n \times p_0 \times (1-p_0) \geq 10 \)[/tex].
- Here, [tex]\( n = 150 \)[/tex] and [tex]\( p_0 = 0.15 \)[/tex].
Calculating:
[tex]\[
n \times p_0 \times (1-p_0) = 150 \times 0.15 \times 0.85 = 19.125
\][/tex]
Since 19.125 is greater than 10, the large counts condition is met.
With these conditions verified, we can proceed with the z-test for one proportion.
### Verification of Statements
- [tex]\(H_0: p=0.15\)[/tex]: True
- [tex]\(H_{a}: p<0.15\)[/tex]: False (the correct alternative hypothesis [tex]\(H_a\)[/tex] is [tex]\(p > 0.15\)[/tex])
- The random condition is met: True
- The 10% condition is met: True
- The large counts condition is met: True
- The test is a [tex]\(z\)[/tex]-test for one proportion: True
Thus, the true statements based on the above analysis are:
- [tex]\(H_0: p=0.15\)[/tex]
- The random condition is met.
- The 10% condition is met.
- The large counts condition is met.
- The test is a [tex]\(z\)[/tex]-test for one proportion.
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