Question 1195769: 1. As part of a study to determine the effect of bumetanide on urinary calcium excretion, nine randomly selected men each received an oral dose of .5 mg of the drug. Urine was collected hourly for the next six hours. The mean excretion rate for this sample of nine men was found to be 7.5 mg/hr, with a standard deviation of 6.0 mg/hr.
a. What is the 98% confidence interval estimate of the population mean excretion rate for all men receiving bumetanide? Interpret this interval.
b. Urine was also collected from a random sample of 16 men who did not receive bumetanide. The mean excretion rate for this sample was 6.5 mg/hr, with a standard deviation of 2.0 mg/hr. What is the 98% confidence interval estimate of the difference in excretion rates between the population of men receiving the drug and the population of men who did not receive it?
Answer by proyaop(69)   (Show Source):
You can put this solution on YOUR website! Certainly, let's calculate the confidence intervals.
**a. 98% Confidence Interval for the Population Mean Excretion Rate (Bumetanide Group)**
* **Given:**
* Sample size (n1) = 9
* Sample mean (x̄1) = 7.5 mg/hr
* Sample standard deviation (s1) = 6.0 mg/hr
* Confidence level = 98%
* **Find:**
* Critical value (tα/2)
* Margin of error (E)
* Confidence interval
1. **Find the critical value (tα/2)**:
* Degrees of freedom (df) = n1 - 1 = 9 - 1 = 8
* α = 1 - 0.98 = 0.02
* α/2 = 0.01
Using a t-distribution table or a calculator, find the t-value for df = 8 and α/2 = 0.01.
* tα/2 ≈ 2.896
2. **Calculate the standard error:**
* Standard error (SE) = s1 / √n1 = 6.0 / √9 = 2.0 mg/hr
3. **Calculate the margin of error (E):**
* E = tα/2 * SE = 2.896 * 2.0 = 5.792 mg/hr
4. **Calculate the confidence interval:**
* Lower limit = x̄1 - E = 7.5 - 5.792 = 1.708 mg/hr
* Upper limit = x̄1 + E = 7.5 + 5.792 = 13.292 mg/hr
* **Interpretation:**
We are 98% confident that the true population mean urinary calcium excretion rate for all men receiving bumetanide lies between 1.708 mg/hr and 13.292 mg/hr.
**b. 98% Confidence Interval for the Difference in Excretion Rates**
* **Given:**
* Sample size for bumetanide group (n1) = 9
* Sample mean for bumetanide group (x̄1) = 7.5 mg/hr
* Sample standard deviation for bumetanide group (s1) = 6.0 mg/hr
* Sample size for no bumetanide group (n2) = 16
* Sample mean for no bumetanide group (x̄2) = 6.5 mg/hr
* Sample standard deviation for no bumetanide group (s2) = 2.0 mg/hr
* Confidence level = 98%
* **Find:**
* Degrees of freedom (df)
* Pooled standard deviation (sp)
* Standard error of the difference (SEdiff)
* Critical value (tα/2)
* Margin of error (E)
* Confidence interval
1. **Calculate degrees of freedom (df):**
* Using the Welch-Satterthwaite equation for unequal variances:
df ≈ [(s1^2/n1 + s2^2/n2)^2] / [(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)]
df ≈ [(6.0^2/9 + 2.0^2/16)^2] / [(6.0^2/9)^2/(9-1) + (2.0^2/16)^2/(16-1)]
df ≈ 14.17
(Round down to 14 for t-distribution table lookup)
2. **Calculate the pooled standard deviation (sp):**
* Since we are assuming unequal variances, we do not calculate a pooled standard deviation.
3. **Calculate the standard error of the difference (SEdiff):**
* SEdiff = √[(s1^2/n1) + (s2^2/n2)]
* SEdiff = √[(6.0^2/9) + (2.0^2/16)]
* SEdiff = √(4 + 0.25) = √4.25 ≈ 2.06 mg/hr
4. **Find the critical value (tα/2):**
* α = 1 - 0.98 = 0.02
* α/2 = 0.01
* Using a t-distribution table with df = 14, find tα/2 for α/2 = 0.01.
* tα/2 ≈ 2.624
5. **Calculate the margin of error (E):**
* E = tα/2 * SEdiff = 2.624 * 2.06 = 5.40 mg/hr
6. **Calculate the confidence interval:**
* Lower limit = (x̄1 - x̄2) - E = (7.5 - 6.5) - 5.40 = -4.40 mg/hr
* Upper limit = (x̄1 - x̄2) + E = (7.5 - 6.5) + 5.40 = 6.40 mg/hr
* **Interpretation:**
We are 98% confident that the true difference in mean urinary calcium excretion rates between the population of men receiving bumetanide and the population of men who did not receive it lies between -4.40 mg/hr and 6.40 mg/hr.
**Note:**
* The confidence intervals are based on the assumption of normality within each group.
* The Welch-Satterthwaite equation is used to approximate the degrees of freedom when the population variances are assumed to be unequal.
I hope this helps! Let me know if you have any further questions.
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