Exploring the role of sample size: a. Construct 1,000 confidence intervals with displaystyle{n}={10},{p}={0.3}. What proportion of the 95% confidence

Step 1
The provided information is:
It is given that:
Sample mean ${\displaystyle \left(\hat{p}\right)=0.3}$
Level of significance ${\displaystyle \left(L\right)=0.05}$
The critical calue of ${\displaystyle z_c}$ at 0.05 level of significance woult be:
${\displaystyle z_c=z_{\frac{L}{2}}}$
${\displaystyle =1.96}$
Step 2
a)
The confidence interval the population proportion (p), at 0.05 level of significance, can be calculated as:
Sample size ${\displaystyle \left(n\right)=10}$
${\displaystyle p\in (\hat{p}\pm z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})}$
${\displaystyle \in (0.3\pm 1.96\times \sqrt{\frac{0.3(1-0.3)}{10}})}$
in (0.0159, 0.5840)
Thus, the confidence interval is (0.0159, 0.5840). Therefore, the population proportion 0.3, lies in the confidence interval.
The confidence interval would be same, 1000 times.
Step 3
b)
Again, the confidence interval for 40 different samples, the population proportion (p), at 0.05 level of significance, can be calculated as:
Sample size ${\displaystyle \left(n\right)=40}$
${\displaystyle p\in (\hat{p}\pm z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})}$
${\displaystyle \in (0.3\pm 1.96\times \sqrt{\frac{0.3(1-0.3)}{40}})}$
in (0.1579, 0.4420)
Thus, the confidence interval is (0.1579, 0.4420). Therefore, the population proportion 0.3, lies in the confidence interval.
The confidence interval would be same, 1000 times.
Step 4
c)
The confidence interval the population proportion (p), at 0.05 level of significance, can be calculated as:
Sample size ${\displaystyle \left(n\right)=100}$
${\displaystyle p\in (\hat{p}\pm z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}})}$
${\displaystyle \in (0.3\pm 1.96\times \sqrt{\frac{0.3(1-0.3)}{100}})}$
${\displaystyle \in (0.2102,0.3898)}$
Thus, the confidence interval is (0.2102, 0.3898). Therefore, the population proportion 0.3, lies in the confidence interval.
The confidence interval would be same, 1000 times.