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KonradZuse
 3 years ago
A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy.
Reference: Ref 183
A 90% confidence interval for p is
A. 0.4489 to 0.5159.
B. 0.4543 to 0.5105.
C. 0.4487 to 0.5161.
D. 0.4463 to 0.5185.
KonradZuse
 3 years ago
A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy. Reference: Ref 183 A 90% confidence interval for p is A. 0.4489 to 0.5159. B. 0.4543 to 0.5105. C. 0.4487 to 0.5161. D. 0.4463 to 0.5185.

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KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @Outkast3r09 @TuringTest

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1let q = sample proportion q = x/n = 410/850 = 0.48235 Now use the formula E = C * sqrt( (q(1q))/n ) to find the margin of error E = C * sqrt( (q(1q))/n ) E = 1.645 * sqrt( ( 0.48235(1 0.48235))/850 ) E = 0.0281939  The 90% confidence interval is then ( q  E , q + E ) ( 0.48235  0.0281939 , 0.48235 + 0.0281939 ) ( 0.4541561, 0.5105439 ) ( 0.4542, 0.5105 ) So it's very close to B, and it's just off because of rounding errors.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0haha I see I messed up I got .43 and wondered why... Thanks Jim as always :D
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