let two stochastically independent random variables y1 and y2 with the distribution b(100,p1) and b(100,p2) respectively,y1=50 and y2=40 ,find 90% a confidence interval for p1-p2 ?
To find a confidence interval for p1 - p2, we need to know the sample sizes n1 and n2, as well as the observed values y1 and y2. In this case, n1 = n2 = 100, y1 = 50, and y2 = 40.
1. Calculate the sample proportions:
p̂1 = y1 / n1 = 50 / 100 = 0.5
p̂2 = y2 / n2 = 40 / 100 = 0.4
2. Calculate the standard errors:
SE1 = sqrt(p̂1 * (1 - p̂1) / n1) = sqrt(0.5 * 0.5 / 100) = 0.05
SE2 = sqrt(p̂2 * (1 - p̂2) / n2) = sqrt(0.4 * 0.6 / 100) = 0.04899 (approximately)
3. Calculate the margin of error:
ME = 1.645 * sqrt(SE1^2 + SE2^2)
= 1.645 * sqrt(0.05^2 + 0.04899^2)
≈ 0.0647 (rounded to four decimal places)
4. Calculate the confidence interval:
CI = (p̂1 - p̂2) ± ME
= (0.5 - 0.4) ± 0.0647
= 0.1 ± 0.0647
= (0.0353, 0.1647)
Therefore, the 90% confidence interval for p1 - p2 is approximately (0.0353, 0.1647).
Note: We assumed a normal approximation because the sample size (n1 = n2 = 100) is large enough for the Central Limit Theorem to apply.