 # 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 snowlovelydayM 2020-12-22 Answered
Exploring the role of sample size:
a. Construct 1,000 confidence intervals with $n=10,p=0.3$. What proportion of the $95\mathrm{%}$ confidence intervals included the population proportion, 0.3?
b. Construct 1,000 confidence intervals with $n=40,p=0.3$. What proportion of the $95\mathrm{%}$ confidence intervals included the population proportion, 0.3?
c. Construct 1,000 confidence intervals with $n=100,p=0.3$. What proportion $95\mathrm{%}$ confidence intervals include the proportion of the population, 0.3?
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Step 1
The provided information is:
It is given that:
Sample mean $\left(\stackrel{^}{p}\right)=0.3$
Level of significance $\left(L\right)=0.05$
The critical calue of ${z}_{c}$ at 0.05 level of significance woult be:
${z}_{c}={z}_{\frac{L}{2}}$
$=1.96$
Step 2
a)
The confidence interval the population proportion (p), at 0.05 level of significance, can be calculated as:
Sample size $\left(n\right)=10$
$p\in \left(\stackrel{^}{p}±{z}_{c}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}\right)$
$\in \left(0.3±1.96×\sqrt{\frac{0.3\left(1-0.3\right)}{10}}\right)$
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 $\left(n\right)=40$
$p\in \left(\stackrel{^}{p}±{z}_{c}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}\right)$
$\in \left(0.3±1.96×\sqrt{\frac{0.3\left(1-0.3\right)}{40}}\right)$
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 $\left(n\right)=100$
$p\in \left(\stackrel{^}{p}±{z}_{c}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}\right)$
$\in \left(0.3±1.96×\sqrt{\frac{0.3\left(1-0.3\right)}{100}}\right)$
$\in \left(0.2102,0.3898\right)$
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.