Brenda takes a simple random sample of 17 professional hockey players and 23 professional baseball players. In her sample, the hockey players had a mean weight of 196.77 pounds with a standard deviation of 16.37 pounds. The baseball players had a mean weight of 210.76 pounds with a standard deviation of 15.43 pounds. Calculate a 95% confidence interval for the differences between weights. Please write your answers to at least four decimal places. a. Critical value: t = b. Degrees of freedom: c. Standard error: d. Point estimate: e. Margin of error: f. Confidence interval: < μα <

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### Determining the 95% Confidence Interval for Differences in Weight Between Professional Hockey and Baseball Players

Brenda takes a simple random sample of 17 professional hockey players and 23 professional baseball players. In her sample, the hockey players had a mean weight of 196.77 pounds with a standard deviation of 16.37 pounds. The baseball players had a mean weight of 210.76 pounds with a standard deviation of 15.43 pounds.

Calculate a 95% confidence interval for the differences between weights. Please write your answers to at least four decimal places.

**Given Data:**

- **Hockey Players** (Group 1):
  - Sample Size (\(n_1\)) = 17
  - Mean (\(\bar{X}_1\)) = 196.77 pounds
  - Standard Deviation (\(S_1\)) = 16.37 pounds

- **Baseball Players** (Group 2):
  - Sample Size (\(n_2\)) = 23
  - Mean (\(\bar{X}_2\)) = 210.76 pounds
  - Standard Deviation (\(S_2\)) = 15.43 pounds
  
**Step-by-Step Solution:**

***a. Critical value: \( t^* \) = *** [This value will depend on the t-distribution table based on the degrees of freedom calculated in the next step.]

***b. Degrees of freedom: ***
\[ \text{Degrees of freedom} \approx \text{Min}(n_1 - 1, n_2 - 1) \]

***c. Standard error: ***
\[ \text{Standard error} = \sqrt{\left( \frac{S_1^2}{n_1} \right) + \left( \frac{S_2^2}{n_2} \right) } \]

***d. Point estimate: ***
\[ \text{Point estimate} = \bar{X}_2 - \bar{X}_1 \]

***e. Margin of error: ***
\[ \text{Margin of error} = t^* \times \text{Standard error} \]

***f. Confidence interval:***
\[ (\text{Point estimate} - \text{Margin of error}) < \mu_d < (\text{Point estimate} + \text{Margin of error}) \]

**Please complete the calculations and
Transcribed Image Text:### Determining the 95% Confidence Interval for Differences in Weight Between Professional Hockey and Baseball Players Brenda takes a simple random sample of 17 professional hockey players and 23 professional baseball players. In her sample, the hockey players had a mean weight of 196.77 pounds with a standard deviation of 16.37 pounds. The baseball players had a mean weight of 210.76 pounds with a standard deviation of 15.43 pounds. Calculate a 95% confidence interval for the differences between weights. Please write your answers to at least four decimal places. **Given Data:** - **Hockey Players** (Group 1): - Sample Size (\(n_1\)) = 17 - Mean (\(\bar{X}_1\)) = 196.77 pounds - Standard Deviation (\(S_1\)) = 16.37 pounds - **Baseball Players** (Group 2): - Sample Size (\(n_2\)) = 23 - Mean (\(\bar{X}_2\)) = 210.76 pounds - Standard Deviation (\(S_2\)) = 15.43 pounds **Step-by-Step Solution:** ***a. Critical value: \( t^* \) = *** [This value will depend on the t-distribution table based on the degrees of freedom calculated in the next step.] ***b. Degrees of freedom: *** \[ \text{Degrees of freedom} \approx \text{Min}(n_1 - 1, n_2 - 1) \] ***c. Standard error: *** \[ \text{Standard error} = \sqrt{\left( \frac{S_1^2}{n_1} \right) + \left( \frac{S_2^2}{n_2} \right) } \] ***d. Point estimate: *** \[ \text{Point estimate} = \bar{X}_2 - \bar{X}_1 \] ***e. Margin of error: *** \[ \text{Margin of error} = t^* \times \text{Standard error} \] ***f. Confidence interval:*** \[ (\text{Point estimate} - \text{Margin of error}) < \mu_d < (\text{Point estimate} + \text{Margin of error}) \] **Please complete the calculations and
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