Question

question 4 suppose that 20% of the residents in a certain state support an increase in the property tax. an opinion poll will randomly sample 400 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 0.04 of the true proportion (i.e., between 0.16 and 0.24)? (hint: use the sampling distribution of the sample proportion in this case.) a. it is certain that the resulting sample proportion will be within 0.04 of the true proportion. b. there is roughly a 99.7% chance that the resulting sample proportion will be within 0.04 of the true proportion. c. there is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion. d. there is roughly a 68% chance that the resulting sample proportion will be within 0.04 of the true proportion.

Explanation:

Step1: Identify the parameters

Let $p = 0.2$ (true - proportion), $n=400$ (sample size). The standard deviation of the sampling distribution of the sample proportion $\hat{p}$ is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$.

Step2: Calculate the standard deviation

Substitute $p = 0.2$ and $n = 400$ into the formula:
$\sigma_{\hat{p}}=\sqrt{\frac{0.2\times(1 - 0.2)}{400}}=\sqrt{\frac{0.2\times0.8}{400}}=\sqrt{\frac{0.16}{400}}=\sqrt{0.0004}=0.02$.

Step3: Calculate the z - scores

The lower limit $L = 0.16$ and the upper limit $U = 0.24$. The z - score for the lower limit $z_1=\frac{0.16 - 0.2}{0.02}=\frac{- 0.04}{0.02}=-2$, and the z - score for the upper limit $z_2=\frac{0.24 - 0.2}{0.02}=\frac{0.04}{0.02}=2$.

Step4: Use the standard normal distribution

We want to find $P(-2

Answer:

C. There is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.