Question

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question is in photo

Answer 1

We need to construct the 98% confidence interval for the population proportion. 

The sample proportion is computed as follows, based on the sample size $N=200$ and the number of favorable cases $X=144:$

$$\hat{p}={X \over N}={144 \over 200}=0.72$$

The critical value for $\alpha=0.02$ is $z_c=z_{1-\alpha/2}=2.326.$ The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=$$

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

$$=\big(0.72-2.326\sqrt{{0.72(1-0.72) \over 200}},0.72+2.326\sqrt{{0.72(1-0.72) \over 200}}\big)\approx$$

$$\approx(0.646,0.794)$$

Therefore, based on the data provided, the 98% confidence interval for the population proportion is $0.646<p<0.794,$ which indicates that we are 98% confident that the true population proportion $p$  is contained by the interval $(0.646,0.794).$

b) The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

$$E=z_c\sqrt{{\hat{p}(1-\hat{p}) \over N}}\leq0.05$$

$${0.72(1-0.72) \over N}\leq({0.05 \over 1.96})^2$$

$$N\geq{0.72(1-0.72)(1.96)^2 \over (0.05)^2}$$

$$N\geq310$$