We appreciate your visit to According to a recent study tex 15 tex of adults who take a certain medication experience side effects To further investigate this finding a researcher. 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 :
Let's go through the steps to determine whether there is convincing statistical evidence that the proportion of adults who experience side effects from the medication is greater than 0.15.
State:
First, we define the null and alternative hypotheses for this test:
- [tex]\( H_0: p = 0.15 \)[/tex]
This null hypothesis states that the true proportion of adults who experience side effects is 0.15.
- [tex]\( H_a: p > 0.15 \)[/tex]
The alternative hypothesis claims that the true proportion of adults who experience side effects is greater than 0.15.
We will be conducting this hypothesis test at a significance level of [tex]\( \alpha = 0.05 \)[/tex].
Plan:
Next, we'll determine if the requirements for conducting a hypothesis test for a proportion are met:
1. Random Condition: This condition requires that the sample is randomly selected. It's stated in the problem that a separate random sample of adults is used, so this condition is satisfied.
2. 10% Condition: We need to ensure the sample size is less than 10% of the population to assume independence. Assuming the population of adults taking this medication is large, 150 is less than 10% of such a population if it's over 1,500. Thus, this condition is met.
3. Large Counts Condition: This ensures that the sample is large enough for the sampling distribution of the sample proportion to be approximately normal:
- [tex]\( n \times p_0 \ge 10 \)[/tex] and [tex]\( n \times (1 - p_0) \ge 10 \)[/tex], where [tex]\( n = 150 \)[/tex] and [tex]\( p_0 = 0.15 \)[/tex].
Calculating these:
- [tex]\( 150 \times 0.15 = 22.5 \)[/tex], which is ≥ 10.
- [tex]\( 150 \times (1 - 0.15) = 150 \times 0.85 = 127.5 \)[/tex], which is ≥ 10.
Therefore, the large counts condition is also satisfied.
The data suggests using the test: z-test for one proportion.
Answering the Given Statements:
- [tex]\( H_0: p = 0.15 \)[/tex]: True
- [tex]\( H_a: p < 0.15 \)[/tex]: False (the correct alternative hypothesis is [tex]\( H_a: p > 0.15 \)[/tex])
- The random condition is met: True
- The 10% condition is met: True
- The large counts condition is met: True
- The test is a z-test for one proportion: True
These statements reflect a setup consistent with conducting a hypothesis test using a z-test for one proportion to evaluate if the proportion of adults experiencing side effects is greater than 0.15.
State:
First, we define the null and alternative hypotheses for this test:
- [tex]\( H_0: p = 0.15 \)[/tex]
This null hypothesis states that the true proportion of adults who experience side effects is 0.15.
- [tex]\( H_a: p > 0.15 \)[/tex]
The alternative hypothesis claims that the true proportion of adults who experience side effects is greater than 0.15.
We will be conducting this hypothesis test at a significance level of [tex]\( \alpha = 0.05 \)[/tex].
Plan:
Next, we'll determine if the requirements for conducting a hypothesis test for a proportion are met:
1. Random Condition: This condition requires that the sample is randomly selected. It's stated in the problem that a separate random sample of adults is used, so this condition is satisfied.
2. 10% Condition: We need to ensure the sample size is less than 10% of the population to assume independence. Assuming the population of adults taking this medication is large, 150 is less than 10% of such a population if it's over 1,500. Thus, this condition is met.
3. Large Counts Condition: This ensures that the sample is large enough for the sampling distribution of the sample proportion to be approximately normal:
- [tex]\( n \times p_0 \ge 10 \)[/tex] and [tex]\( n \times (1 - p_0) \ge 10 \)[/tex], where [tex]\( n = 150 \)[/tex] and [tex]\( p_0 = 0.15 \)[/tex].
Calculating these:
- [tex]\( 150 \times 0.15 = 22.5 \)[/tex], which is ≥ 10.
- [tex]\( 150 \times (1 - 0.15) = 150 \times 0.85 = 127.5 \)[/tex], which is ≥ 10.
Therefore, the large counts condition is also satisfied.
The data suggests using the test: z-test for one proportion.
Answering the Given Statements:
- [tex]\( H_0: p = 0.15 \)[/tex]: True
- [tex]\( H_a: p < 0.15 \)[/tex]: False (the correct alternative hypothesis is [tex]\( H_a: p > 0.15 \)[/tex])
- The random condition is met: True
- The 10% condition is met: True
- The large counts condition is met: True
- The test is a z-test for one proportion: True
These statements reflect a setup consistent with conducting a hypothesis test using a z-test for one proportion to evaluate if the proportion of adults experiencing side effects is greater than 0.15.
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