ie hoogte van ‘n ewekansige van 50 Hereford kalwers by geboorte toon ‘n gemiddelde van 71 cm en ‘n standaardafwyking van 6.7 cm. Konstrueer ‘n 90% vertrouensinterval vir gemiddelde hoogte van alle Hereford kalwers by geboorte. / 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.


[68.5592 ; 73.4408]


[69.4414 ; 72.5586]


[69.1429 ; 72.8571]


[69.4414 ; 73.4408]

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 population standard deviation is unknown. This is calculated using the t-distribution.

The formula for the confidence interval is given by:

\[ \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \]

Where:

  • \(\bar{x}\) is the sample mean
  • \(t_{\alpha/2}\) is the t-critical value for \(n-1\) degrees of freedom at the desired confidence level
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size

Given:

  • Sample size \(n = 50\)
  • Sample mean \(\bar{x} = 71\) cm
  • Sample standard deviation \(s = 6.7\) cm

Step 1: Determine the t-critical value

For a 90% confidence interval and \(n - 1 = 50 - 1 = 49\) degrees of freedom, you can find the t-critical value using t-distribution tables or software. For 90% confidence, \(\alpha = 0.10\), so \(\alpha/2 = 0.05\).

Looking at the t-table, the t-critical value for 49 degrees of freedom is approximately \(t_{0.05, 49} \approx 1.676\).

Step 2: Calculate the margin of error (ME)

\[ ME = t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) = 1.676 \left(\frac{6.7}{\sqrt{50}}\right) \]

Calculating \(\frac{s}{\sqrt{n}}\): \[ \frac{6.7}{\sqrt{50}} \approx \frac{6.7}{7.071} \approx 0.9487 \]

Now compute the margin of error: \[ ME \approx 1.676 \times 0.9487 \approx 1.5887 \]

Step 3: Construct the confidence interval

Now use the mean and margin of error to construct the confidence interval: \[ CI = \bar{x} \pm ME = 71 \pm 1.5887 \]

Calculating:

  • Lower limit: \(71 - 1.5887 \approx 69.4113\)
  • Upper limit: \(71 + 1.5887 \approx 72.5887\)

Final Result

So the 90% confidence interval for the mean height of all Hereford calves at birth is approximately: \[ [69.4113; 72.5887] \]

Now comparing this with the provided options:

  • [68.5592 ; 73.4408]
  • [69.4414 ; 72.5586]
  • [69.1429 ; 72.8571]
  • [69.4414 ; 73.4408]

The closest to our calculated interval is [69.4414 ; 72.5586]. Thus, the answer is:

[69.4414 ; 72.5586].