Question

Two years ago, a political party ASR received 9.8% of the votes in an election. To study the current political preferences, a statistical research institute plans to organise a poll by the end of the present year. In this study, n voters will be interviewed about the political party they prefer. Below, p, denotes the proportion of voters that would vote ASR if the elections were held now. Furthermore p^ denotes the (random) sample proportion of the ASR voters.d) At the end of the year, n randomly chosen voters are interviewed with n as calculated in part(c); the realisation of p^turns out to be 0.853.Use the interval in part (b) as a starting point to create an interval that will probably contain the population proportion p. Is the width of that interval indeed about 0.02?

Answer 1

We need to construct the $90\%$ confidence interval for the population proportion. We have been provided with the following information about the sample proportion:

$\begin{matrix} Sample\ proportion & \hat{p}=0.098 \\ Sample\ Size & N=500 \end{matrix}$

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.645.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=$$

$$=\big(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}},\hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}} \big)$$

$$=(0.098-1.645\sqrt{{0.098(1-0.098) \over 500}},0.098+1.645\sqrt{{0.098(1-0.098) \over 500}})=$$

$$=(0.076,0.120)$$

How large should the sample size be to Obtain an interval width about 0:02?

An interval width is

$$width=\hat{p}+z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}-(\hat{p}-z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}})=$$

$$=2z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}$$

$$2(1.645)\sqrt{{0.098(1-0.098) \over N}}=0.02$$

$$N={ 0.098(1-0.098)\over \big(\dfrac{0.02}{2(1.645)}\big)^2}\approx2392$$