Question 1195769
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.