Question

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 900 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion to be within .02 of the true proportion (i.e., between .18 and .22)? (Hint: Use the sampling distribution of the sample proportion in this case.)

Answer 1

86.64% probability that the resulting sample proportion is within .02 of the true proportion.

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

In this problem:

`\mu = 0.2, \sigma = \sqrt{(0.2*0.8)/(900)} = 0.0133`

How likely is the resulting sample proportion to be within .02 of the true proportion (i.e., between .18 and .22)?

This is the pvalue of Z when X = 0.22 subtracted by the pvalue of Z when X = 0.18.

X = 0.22

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

`Z = (0.22 - 0.2)/(0.0133)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332.

X = 0.18

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

`Z = (0.18 - 0.2)/(0.0133)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that the resulting sample proportion is within .02 of the true proportion.