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Answer :
To address the question of whether the new diabetes medication results in less variance in a patient's glucose level compared to no medication, we will undertake a hypothesis test. Here is the detailed step-by-step solution:
### Step 1: State the null and alternative hypotheses.
Null Hypothesis (H0): The variance of glucose levels for the medication group (Group A) is not less than that for the placebo group (Group B). Symbolically, this can be expressed as:
[tex]\[ H_0: \frac{\sigma_A^2}{\sigma_B^2} \geq 1 \][/tex]
Alternative Hypothesis (Ha): The variance of glucose levels for the medication group (Group A) is less than for the placebo group (Group B). This can be written as:
[tex]\[ H_a: \frac{\sigma_A^2}{\sigma_B^2} < 1 \][/tex]
This indicates we are performing a left-tailed test.
### Step 2: Identify the sampling distribution.
When comparing variances, the F-distribution is used. The test statistic, called the F-statistic, is calculated as the ratio of the sample variances:
[tex]\[ F = \frac{S_A^2}{S_B^2} \][/tex]
where [tex]\(S_A^2\)[/tex] and [tex]\(S_B^2\)[/tex] are the sample variances of Group A and Group B, respectively.
Let's list the degrees of freedom:
- Degrees of freedom for Group A: [tex]\( n_A - 1 = 26 - 1 = 25 \)[/tex]
- Degrees of freedom for Group B: [tex]\( n_B - 1 = 31 - 1 = 30 \)[/tex]
### Step 3: Calculate the sample variances.
From the data, we have:
- The variance for Group A ([tex]\(S_A^2\)[/tex]) is approximately 1757.22.
- The variance for Group B ([tex]\(S_B^2\)[/tex]) is approximately 1667.73.
### Step 4: Calculate the F-statistic.
Using the variances:
[tex]\[ F = \frac{1757.22}{1667.73} \approx 1.0537 \][/tex]
### Step 5: Find the p-value for the test.
The p-value indicates the probability of observing a test statistic as extreme as, or more extreme than, the one computed, assuming the null hypothesis is true. For a left-tailed test, the p-value can be calculated from the cumulative distribution function (CDF) of the F-distribution.
The calculated p-value is approximately 0.5584.
### Step 6: Make a decision based on the p-value.
At a 5% significance level ([tex]\(\alpha = 0.05\)[/tex]):
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
Since 0.5584 is greater than 0.05, we fail to reject the null hypothesis.
### Conclusion:
There is not enough evidence to support the claim that the new diabetes medication results in a smaller variance in glucose levels compared to not taking the medication. Therefore, the pharmaceutical company's claim cannot be substantiated based on this data.
### Step 1: State the null and alternative hypotheses.
Null Hypothesis (H0): The variance of glucose levels for the medication group (Group A) is not less than that for the placebo group (Group B). Symbolically, this can be expressed as:
[tex]\[ H_0: \frac{\sigma_A^2}{\sigma_B^2} \geq 1 \][/tex]
Alternative Hypothesis (Ha): The variance of glucose levels for the medication group (Group A) is less than for the placebo group (Group B). This can be written as:
[tex]\[ H_a: \frac{\sigma_A^2}{\sigma_B^2} < 1 \][/tex]
This indicates we are performing a left-tailed test.
### Step 2: Identify the sampling distribution.
When comparing variances, the F-distribution is used. The test statistic, called the F-statistic, is calculated as the ratio of the sample variances:
[tex]\[ F = \frac{S_A^2}{S_B^2} \][/tex]
where [tex]\(S_A^2\)[/tex] and [tex]\(S_B^2\)[/tex] are the sample variances of Group A and Group B, respectively.
Let's list the degrees of freedom:
- Degrees of freedom for Group A: [tex]\( n_A - 1 = 26 - 1 = 25 \)[/tex]
- Degrees of freedom for Group B: [tex]\( n_B - 1 = 31 - 1 = 30 \)[/tex]
### Step 3: Calculate the sample variances.
From the data, we have:
- The variance for Group A ([tex]\(S_A^2\)[/tex]) is approximately 1757.22.
- The variance for Group B ([tex]\(S_B^2\)[/tex]) is approximately 1667.73.
### Step 4: Calculate the F-statistic.
Using the variances:
[tex]\[ F = \frac{1757.22}{1667.73} \approx 1.0537 \][/tex]
### Step 5: Find the p-value for the test.
The p-value indicates the probability of observing a test statistic as extreme as, or more extreme than, the one computed, assuming the null hypothesis is true. For a left-tailed test, the p-value can be calculated from the cumulative distribution function (CDF) of the F-distribution.
The calculated p-value is approximately 0.5584.
### Step 6: Make a decision based on the p-value.
At a 5% significance level ([tex]\(\alpha = 0.05\)[/tex]):
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
Since 0.5584 is greater than 0.05, we fail to reject the null hypothesis.
### Conclusion:
There is not enough evidence to support the claim that the new diabetes medication results in a smaller variance in glucose levels compared to not taking the medication. Therefore, the pharmaceutical company's claim cannot be substantiated based on this data.
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