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 <
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 <
MATLAB: An Introduction with Applications
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8e28faf-547f-4c48-bb6d-e40546588d1c%2F51c8419e-a392-4f4d-bbf2-9b7d38016e1b%2Flzmw2on_processed.jpeg&w=3840&q=75)
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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