To construct a 99% confidence interval for the mean height of all Hereford calves at birth, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown, which in this case we will approximate using the sample standard deviation.
The formula for the confidence interval (CI) is:
\[ \text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(t^*\) is the t-value for a 99% confidence level with \(n - 1\) degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
Given:
- Sample mean (\(\bar{x}\)) = 71 cm
- Sample standard deviation (\(s\)) = 6.7 cm
- Sample size (\(n\)) = 50
First, we find the critical t-value (\(t^\)) for a 99% confidence level with \(n - 1 = 50 - 1 = 49\) degrees of freedom. You can find this value in a t-distribution table or using statistical software. For 49 degrees of freedom at a 99% confidence level, the \(t^\) value is approximately 2.676.
Next, we calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{6.7}{\sqrt{50}} \approx \frac{6.7}{7.071} \approx 0.948 \]
Now we can calculate the margin of error (ME):
\[ ME = t^* \times SE = 2.676 \times 0.948 \approx 2.53 \]
Finally, we can construct the confidence interval:
\[ \text{CI} = (\bar{x} - ME, \bar{x} + ME) = (71 - 2.53, 71 + 2.53) = (68.47, 73.53) \]
Thus, the 99% confidence interval for the mean height of all Hereford calves at birth is approximately:
\[ (68.47 \text{ cm}, 73.53 \text{ cm}) \]