Answer to Question #107317 in Statistics and Probability for NKULULEKO HAROLD LOUW

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..b)Find random bounds (depending on p) that will include the proportion p with probability 0.95.Express the width of the accompanying interval in terms of p and n.c)The interval in part (b) will be the starting point to create, at the end of the current year, when the data are observed, an interval that will probably contain the proportion p. How large should the sample size be to obtain an interval width about 0.02? (Hint: Substitute the former proportion 0.098 for p in the width of part (b).

Answer 1

We need to construct the $95\%$ 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 $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=$$

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

$$=(0.098-1.96\sqrt{{0.098(1-0.098) \over 500}},0.098+1.96\sqrt{{0.098(1-0.098) \over 500}})=$$

$$=(0.072,0.124)$$

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

An interval width is

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

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

$$2(1.96)\sqrt{{0.098(1-0.098) \over N}}=0.02$$

$$N={ 0.098(1-0.098)\over \big(\dfrac{0.02}{2(1.96)}\big)^2}\approx3396$$