Exploring the role of sample size: a. Construct 1,000 confidence intervals with n=10,p=0.3. What proportion of the 95% 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% confidence intervals included the population proportion, 0.3? c. Construct 1,000 confidence intervals with n=100,p=0.3. What proportion 95% confidence intervals include the proportion of the population, 0.3?

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Exploring the role of sample size:  a. Construct 1,000 confidence intervals with n=10,p=0.3. What proportion of the 95% 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% confidence intervals included the population proportion, 0.3?  c. Construct 1,000 confidence intervals with n=100,p=0.3. What proportion 95% confidence intervals include the proportion of the population, 0.3?

ensojadasH · Accepted Answer

Step 1  The provided information is:  It is given that:  Sample mean (p^)=0.3  Level of significance (L)=0.05  The critical calue of zc at 0.05 level of significance woult be:  zc=zL2  =1.96  Step 2  a)  The confidence interval the population proportion (p), at 0.05 level of significance, can be calculated as:  Sample size (n)=10  p∈(p^±zcp^(1−p^)n)  ∈(0.3±1.96×0.3(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 (n)=40  p∈(p^±zcp^(1−p^)n)  ∈(0.3±1.96×0.3(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 (n)=100  p∈(p^±zcp^(1−p^)n)  ∈(0.3±1.96×0.3(1−0.3)100)  ∈(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.

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

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