Question

According to a survey, 63% of the Scottish population has visited woodland in the previous year. Skeptical about this claim, you decide to take a simple random sample of 650 people in this population, and you ask them if they had visited woodland in the previous year. You find that 60% of the sample replied "yes" to your question.

Assuming that the original survey's 63% claim is correct, what is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?

Answer 1

0.06

Explanation:

khan

Answer 2

5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a sampling propotion p in a sample of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`\mu = 0.63, \sigma = \sqrt{(0.63*0.37)/(650)} = 0.0189`

What is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?

This is the pvalue of Z when X = 0.6. So

`Z = (X - \mu)/(\sigma)`

`Z = (0.6 - 0.63)/(0.0189)`

`Z = -1.58`

`Z = -1.58` has a pvalue of 0.0571

5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year