A newsgroup is interested in constructing a 95% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 556 randomly selected Americans surveyed, 421 were in favor of the initiative. Round answers to 4 decimal places where possible. Required:
a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between and .

b. If many groups of 593 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.

(A) With 99% confidence, the proportion that favours the Green initiative is between 418.895 and 423.105 (no approximation; answers were gotten exactly in 3 decimal places).

(B) If many groups of 593 (greater than the initial sample size of 556) randomly selected Americans were surveyed then a different confidence interval would be used or produced from each group.

About 90% of these confidence intervals will contain the true population proportion of Americans who favour the Green initiative and about 10% will not contain the true population proportion.

Step-by-step explanation:

(A) 99% of 421 = 416.79

421 - 416.79 = 4.21

4.21 ÷ 2 = 2.105

(421-2.105), (421+2.105)

The lower and upper limits are:

[418.895 , 423.105]

(B) A wider confidence interval such as 90% will be better suited in a case of multiple samples like this.

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step-by-step explanation: