Ans:
Margin of error=0.03
confidence level=0.94
alpha=1-0.94=0.06
critical z value=normsinv(0.03) or normsinv(0.97)=1.88
a)
When prior estimate=0.48
Sample size,n=1.88^2*0.48*(1-0.48)/0.03^2=980
*(if exact z value is considered,then n=981)
b)
when no prior estimate of p,then use p=0.5
Sample size,n=1.88^2*0.5*(1-0.5)/0.03^2=982
*(if exact z value is considered,then n=983)
c)
Option A is correct.
Results are close,because 0.48*(1-0.48)=0.2496 is very close to 0.25
1=9
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Ans:
Margin of error=0.03
confidence level=0.94
alpha=1-0.94=0.06
critical z value=normsinv(0.03) or normsinv(0.97)=1.88
a)
When prior estimate=0.48
Sample size,n=1.88^2*0.48*(1-0.48)/0.03^2=980
*(if exact z value is considered,then n=981)
b)
when no prior estimate of p,then use p=0.5
Sample size,n=1.88^2*0.5*(1-0.5)/0.03^2=982
*(if exact z value is considered,then n=983)
c)
Option A is correct.
Results are close,because 0.48*(1-0.48)=0.2496 is very close to 0.25
2 of 5 (3 complete) A television sports commentator wants to estimate the proportion of citizens...
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Homework: HW 9.1 9.2 9.3 Score: 0 of 1 pt 17 of 40 (25 complete) 9.1.39 A television sports commentator wants to estimate the proportion of citizens who follow professional football." Complete parts (a) through (c). (a) What sample size should be obtained if he wants to be within 4 percentage points with 95% confidence if he uses an estimate of 48% obtained from a poll? The sample size is 600. (Round up to the nearest integer.) (b) What sample...