Question

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 559 randomly selected Americans surveyed, 370 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 506 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.

Answer 1

Answer:

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between 0.6290 and 0.6948.

b. If the sample size is changed, the confidence interval changes as the standard error depends on sample size.

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.

Step-by-step explanation:

We have to calculate a 90% confidence interval for the proportion.

The sample proportion is p=0.6619.

$p=X/n=370/559=0.6619$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.6619*0.3381}{559}}\\\\\\ \sigma_p=\sqrt{0.0004}=0.02$

The critical z-value for a 90% confidence interval is z=1.6449.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.6449 \cdot 0.02=0.0329$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.6619-0.0329=0.6290\\\\UL=p+z \cdot \sigma_p = 0.6619+0.0329=0.6948$

The 90% confidence interval for the population proportion is (0.6290, 0.6948).