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

remolatg 2020-11-16 Answered
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 555 randomly selected Americans surveyed, 437 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 555 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.
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Expert Answer

comentezq
Answered 2020-11-17 Author has 106 answers
Step 1
Consider,
Number of Americans surveyed (n)=555
Number of Americans that are in favor of the initiative (x)=437
(a)
The sample proportion is computed as:
Sample proportion (p^)=xn
=437555
=0.7874
Step 2
From standard normal table, the z-score at 90% confidence level is 1.645.
The 90% confidence interval is computed as:
CI=p^±zα/2×p^(1p^)n
=0.7874±1.645×(0.7874)(10.7874)555
=0.7874±0.0286
=(0.7588,0.8160)
Therefore, the proportion of Americans that favor the new Green Initiative is between 0.7588 and 0.8160.
Step 3
(b)
The confidence interval provides the range in which the population proportion is likely to fall in. In this case, there is 90% probability that the proportion of Americans that favor the new Green initiative is between 0.7588 and 0.8160.
If many groups of 555 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.
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