91Ó°ÊÓ

As per the report published by SANKO, 80% of European dairy farms carry out calf dehorning.

As per the Journal of Dairy Science (Vol. 94, 2011), out of 639 Italian dairy farms, 515 dehorned calves.

That is

The size of the sample is$n=639$

The sample proportion is

$$\begin{gathered} \hat{p}=\frac{515}{639} \\ =0.806 \end{gathered}$$

Where$\hat{p}$ is the sample proportion of Italian dairy farms out calf dehorning.

We have to test the claim reported by SANKO that 80% of European dairy farms carry out calf dehorning by using the survey published in the Journal of Dairy Science.

The null and alternative hypotheses are given as

$H_0:p=0.80$

That is, the true proportion of Italian dairy farms that carry out calf dehorning is 80%.

And

$H_a:p≠0.80$

That is, the proportion of Italian dairy farms carrying out calf dehorning is different from

The test statistic for testing these hypotheses is

$$\begin{gathered} Z=\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} \\ =\frac{0.806-0.80}{\sqrt{\frac{0.80\left(1-0.80\right)}{639}}} \\ =\frac{0.006}{\sqrt{0.0002504}} \\ =0.38 \end{gathered}$$

Assume

$\mathrm{Î\pm }:$The level of significance (chance of making a type I error)

$\mathrm{Î\pm }=.05$

Using the standard normal table, the critical value at the 5% significance level is 1.96.

We can see that

$Z=0.38<1.96$

Hence, we failed to reject the null hypothesis.

At a 5% significance level, we have sufficient evidence to support the figure reported by SANKO that 80% of European dairy farms carry out calf dehorning.