The heights of a random sample of 50 Hereford calves at birth showed a mean of 71 cm and a standard deviation of 6.7 cm. Construct a 90% confidence interval for the mean height of all Hereford calves at birth.

1 answer

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:

\[ (69.44 , \text{cm}, 72.56 , \text{cm}) \]