A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 536 randomly selected Americans surveyed, 441 were in favor of the initiative. Round answers to 4 decimal places where possible.
a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between and .
b. If many groups of 536 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about percent will not contain the true population proportion.
Step 1
Here, sample size ${\displaystyle \left(n\right)=536}$
The sample proportion of Americans who favor the initiative is:
${\displaystyle wide\hat{p}=\frac{411}{536}}$
${\displaystyle =0,8228}$
Step 2 Using the z-table, the two-sided critical value at 90% confidence level is 1.645.
Now, the 90% confidence interval for proportion of all Americans who are in favor of a new Green initiative is calculated as follows:
${\displaystyle wide\hat{p}\pm z_{\frac{\alpha }{2}}\times \sqrt{\frac{0.8228(1-0.8228)}{536}}}$
${\displaystyle 0.8228\pm 1.645\times 0.0165}$
${\displaystyle 0.8228\pm 0.0271}$
(0.7957,0.8499)
Step 3
The 90% confidence interval for population proportion can be interpreted as follows:
There is 90% confidence that the population proportion lie between the lower and upper limit of the calculated interval.
If many samples of size n is obtained from the given population and confidence intervals were calculated for each sample, then about 90% of these intervals contain the true population proportion and the rest 10% will not contain the true population proportion.
a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between 0.7957 and 0.8499.
b. If many groups of 536 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 90% percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 10% percent will not contain the true population proportion.