To construct a 90% 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 sample size is large (n > 30):
\[ \text{CI} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right) \]
where:
- \(\bar{x}\) is the sample mean,
- \(z^*\) is the z-score corresponding to the desired confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.
Step 1: Identify the values
- Sample mean, \(\bar{x} = 71 , \text{cm}\)
- Sample standard deviation, \(s = 6.7 , \text{cm}\)
- Sample size, \(n = 50\)
Step 2: Find the z-score for a 90% confidence level
For a 90% confidence interval, the alpha level (\(\alpha\)) is 0.10. This means that there is 5% in each tail (0.05) of the normal distribution. The corresponding z-score can be found in z-tables or using statistical software:
\[ z^* \approx 1.645 \]
Step 3: Calculate the standard error (SE)
The standard error (SE) is calculated as:
\[ SE = \frac{s}{\sqrt{n}} = \frac{6.7}{\sqrt{50}} \approx \frac{6.7}{7.071} \approx 0.9487 \]
Step 4: Calculate the margin of error (ME)
The margin of error (ME) is given by:
\[ ME = z^* \times SE = 1.645 \times 0.9487 \approx 1.558 \]
Step 5: Construct the confidence interval
Now, we can construct the confidence interval:
\[ \text{CI} = \bar{x} \pm ME = 71 \pm 1.558 \]
Calculating the lower and upper limits:
- Lower limit: \(71 - 1.558 \approx 69.442\)
- Upper limit: \(71 + 1.558 \approx 72.558\)
Final Result
The 90% confidence interval for the mean height of all Hereford calves at birth is approximately:
\[ (69.44 , \text{cm}, 72.56 , \text{cm}) \]
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