To find the 90% confidence interval for the true mean weight of all college wrestlers, we can use the formula for the confidence interval:
\[ \text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean.
- \(t^*\) is the critical value from the t-distribution for the desired confidence level.
- \(s\) is the sample standard deviation.
- \(n\) is the sample size.
Given:
- \(\bar{x} = 276\) pounds
- \(s = 12\) pounds
- \(n = 6\)
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Find the critical value \(t^*\) for a 90% confidence interval with \(n - 1 = 6 - 1 = 5\) degrees of freedom. Using a t-table, the value for \(t^*\) at 90% confidence and 5 degrees of freedom is approximately 2.015.
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Calculate the margin of error (ME): \[ \text{ME} = t^* \left( \frac{s}{\sqrt{n}} \right) = 2.015 \left( \frac{12}{\sqrt{6}} \right) \] First, calculate \(\frac{s}{\sqrt{n}}\): \[ \frac{s}{\sqrt{n}} = \frac{12}{\sqrt{6}} \approx 4.89898 \] Now calculate the margin of error: \[ \text{ME} \approx 2.015 \times 4.89898 \approx 9.877 \]
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Calculate the confidence interval: \[ \text{CI} = 276 \pm 9.877 \] This gives us: \[ (276 - 9.877, 276 + 9.877) = (266.123, 285.877) \]
Rounding to the nearest whole number, we have: \[ (266, 286) \]
So, the 90% confidence interval for the true mean weight of all college wrestlers is \((266,286)\).
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