We have studied the effect of the sample size on the margin of error of the conf
ID: 3060024 • Letter: W
Question
We have studied the effect of the sample size on the margin of error of the confidence interval for a single proportion. In this exercise we perform some calculations to observe this effect for the two-sample problem. Suppose that p1 = 0.7 and p2 = 0.5, and n represents the common value of n1 and n2. Compute the 95% margins of error for the difference between the two proportions for n = 60, 70, 80, 100, 400, 500, and 1000. Present the results in a table. (Give the large-sample margins of error. Round your answers to three decimal places.) n m 60 Incorrect: Your answer is incorrect. 70 80 100 400 500 1000 Present the results with a graph. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot Write a short summary of your findings. As sample size increases, margin of error remains constant. As sample size increases, margin of error increases. As sample size increases, margin of error decreases. There is not enough information.
Explanation / Answer
a) For n = 60
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= 0.7 * 60 + 0.5 * 60)/120 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/60 + 1/60))
= 0.0894
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0984 = 0.193
b) n = 70
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= 0.7 * 70 + 0.5 * 70)/140 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/70 + 1/70))
= 0.0828
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0828 = 0.162
c) For n = 80
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.7 * 80 + 0.5 * 80)/160 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/80 + 1/80))
= 0.0775
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0775 = 0.152
d) For n = 100
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.7 * 100 + 0.5 * 100)/200 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/100 + 1/100))
= 0.0693
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0693 = 0.136
e) For n = 400
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.7 * 400 + 0.5 * 400)/800 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/400 + 1/400))
= 0.0346
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0346 = 0.068
f) For n = 500
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.7 * 500 + 0.5 * 500)/1000 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/500 + 1/500))
= 0.031
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.031 = 0.061
g) For n = 1000
Pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)
= (0.7 * 1000 + 0.5 * 1000)/2000 = 0.6
SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))
= sqrt(0.6 * 0.4 * (1/1000 + 1/1000))
= 0.0219
At 95% confidence interval margin of error is
z0.025 * SE
= 1.96 * 0.0219 = 0.043
As sample size increases margin of error decreases.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.