95% CI for Proportion is p +/- E; where E = Z*sqrt(p*q/n). P=.6, q=.4, n=144, Z=1.96. Therefore E=1.96*sqrt(0.4*0.6/144) = 0.08. So, 95% CI = .6 +/- 0.08 or .52 to .68.
The Z-value for a one-sided 91% confidence interval is 1.34
It depends whether or not the observations are independent and on the distribution of the variable that is being measured or the sample size. You cannot simply assume that the observations are independent and that the distribution is Gaussian (Normal).
Never!
1.96
1.15
There is a 95% probability that the true population proportion lies within the confidence interval.
No, it is not. A 99% confidence interval would be wider. Best regards, NS
The Z-value for a one-sided 91% confidence interval is 1.34
decrease
It depends whether or not the observations are independent and on the distribution of the variable that is being measured or the sample size. You cannot simply assume that the observations are independent and that the distribution is Gaussian (Normal).
4.04%
For a two-tailed interval, they are -1.645 to 1.645
Never!
Estimated p = 75 / 250 = 0.3 Variance of proportion = p*(1-p)/n = 0.3(0.7)/250 =0.00084 S.D. of p is sqrt[0.00084] = 0.029 Confidence interval: phat-zval*sd = 0.3 - (1.96)(0.028983) phat-zval*sd = 0.3 + (1.96)(0.028983) Confidence interval is ( 0.2432 , 0.3568 )
1.96
1.15
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