Question

The united kingdom forestry commission reports that 43\%43%43, percent of the 3.163.163, point, 16 million hectares of woodland area in the united kingdom had certification identifying them as "sustainably managed" in 2016. suppose an employee took a simple random sample of 400400400 of the hectares and saw that the records showed that 47\%47%47, percent of the sampled hectares had that certification in 2016. assuming that the forestry commission's report is accurate, what is the approximate probability that more than 47\%47%47, percent of the sample would have had the certification in 2016?

Answer 1

We can approach this problem using the normal distribution since we have a large sample size and can assume that the sampling distribution of the sample proportion is approximately normal.

First, we need to find the standard error of the sample proportion:

SE = sqrt[(p-hat * (1 - p-hat))/n]
SE = sqrt[(0.47 * 0.53)/400]
SE = 0.030

Next, we can find the z-score:

z = (x - mu) / SE
z = (0.47 - 0.43) / 0.030
z = 1.33

Using a standard normal distribution table, we can find the probability of obtaining a z-score greater than 1.33:

P(z > 1.33) = 0.0918

Therefore, the approximate probability that more than 47% of the sample would have had the certification in 2016 is 0.0918, or about 9.18%.