# 8.2.2.1.3 - Example: Milk

A study of 66,831 dairy cows found that the mean milk yield was 12.5 kg per milking with a standard deviation of 4.3 kg per milking (data from Berry, et al., 2013). Construct a 95% confidence interval for the average milk yield in the population.

First, let's compute the standard error:

$$SE=\frac{s}{\sqrt{n}}=\frac{4.3}{\sqrt{66831}}=0.0166$$

The standard error is small because the sample size is very large.

Next, let's find the $$t^*$$ multiplier:

$$df=66831-1=66830$$

$$t^{*}=1.960$$

Now, we can construct our 95% confidence interval:

95% C.I.: $$12.5\pm1.960(0.017)=12.5\pm0.033=[12.467,\;12.533]$$

We are 95% confident that the mean milk yield in the population is between 12.467 and 12.533 kg per milking.

Berry, D. P., Coyne, J., Boughlan, B., Burke, M., McCarthy, J., Enright, B., Cromie, A. R., McParland, S. (2013). Genetics of milking characteristics in dairy cows. Animal, 7(11), 1750-1758.