Suppose a bag of 122 M&M\'s has 21 orange ones. (A) construct 95% confidence int

Question

Suppose a bag of 122 M&M's has 21 orange ones.
(A) construct 95% confidence interval for the proportion of orange m&m's per bag
(B) what is the margin of error for the proportion of orange m&m's per bag.
(C) is the margin of error a good approximation for the half-width of the confidence interval?
Suppose a bag of 122 M&M's has 21 orange ones.
(A) construct 95% confidence interval for the proportion of orange m&m's per bag
(B) what is the margin of error for the proportion of orange m&m's per bag.
(C) is the margin of error a good approximation for the half-width of the confidence interval?

(A) construct 95% confidence interval for the proportion of orange m&m's per bag
(B) what is the margin of error for the proportion of orange m&m's per bag.
(C) is the margin of error a good approximation for the half-width of the confidence interval?

Explanation / Answer

(A) construct 95% confidence interval for the proportion of orange m&m's per bag

P=21/122 =

Z value for 95% confidence = 1.96

Sample Proportion = 21/122=

0.172

Z Value

1.96

Standard Error of the Proportion = =sqrt( p*(1-p)/n)) = sqrt( 0.172*( 1-0.172)/122))=

0.0342

Margin of error = 1.96*0.0342=

0.0670

Confidence Interval

Interval Lower Limit

0.105

Interval Upper Limit

0.239

(B) what is the margin of error for the proportion of orange m&m's per bag.

Margin of error = 1.96*0.0342=

0.0670

(C) is the margin of error a good approximation for the half-width of the confidence interval?

Yes, the margin of error is a good approximation for the half-width of the confidence interval.

In this case np and n(1-p) > 5.

Sample Proportion = 21/122=

0.172

Z Value

1.96

Standard Error of the Proportion = =sqrt( p*(1-p)/n)) = sqrt( 0.172*( 1-0.172)/122))=

0.0342

Margin of error = 1.96*0.0342=

0.0670

Confidence Interval

Interval Lower Limit

0.105

Interval Upper Limit

0.239

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