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The table to the right shows the results of a survey in which 2552 adults from...

The table to the right shows the results of a survey in which 2552 adults from Country​ A, 1146 adults from Country​ B, and 1066 adults from Country C were asked if human activity contributes to global warming. Complete parts​ (a), (b), and​ (c).

Adults who say that human activity contributes to global warming

Country Upper A 60​%

Country Upper B 89​%

Country Upper C 93​%

​(a) Construct a 95​% confidence interval for the proportion of adults from Country Upper A who say human activity contributes to global warming. left parenthesis nothing comma nothing right parenthesis ​(Round to three decimal places as​ needed.)

(b) Construct a 90​% confidence interval for the proportion of adults from Country Upper B

who say that human activity contributes to global warming.

​(c) Construct a 90% confidence interval for the proportion of adults from Country Upper C who say that human activity contributes to global warming.

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Answer #1

a) For Country A

n = 2552, p̄ = 0.6

95% Confidence interval for the proportion of adults from Country A who say human activity contributes to global warming:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.6 - 1.96 *√(0.6*0.4/2552) = 0.5810

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.6 + 1.96 *√(0.6*0.4/2552) = 0.6190

(0.581, 0.619)

------------------------------------

b) For Country B:

n = 1146, p̄ = 0.89

90% Confidence interval for the proportion of adults from Country B:

At α = 0.10, two tailed critical value, z_c = NORM.S.INV(0.10/2) = 1.645

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.89 - 1.645 *√(0.89*0.11/1146) = 0.8748

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.89 + 1.645 *√(0.89*0.11/1146) = 0.9052

(0.875, 0.905)

------------------------------------

c) For Country C:

n = 1066p̄ = x/n = 0.93

90% Confidence interval for the proportion of adults from Country C:

At α = 0.10, two tailed critical value, z_c = NORM.S.INV(0.10/2) = 1.645

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.93 - 1.645 *√(0.93*0.07/1066) = 0.9171

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.93 + 1.645 *√(0.93*0.07/1066) = 0.9429

(0.917, 0.943)

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