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anyone help on this?
In a simple random sample of 1200 young people, 87% had earned a high school diploma.
a. what is the standard error for this estimate of the percentage of all young people who earned a high school diploma ?
b. find the margin of error using a 95% confidence level for estimating the percentage of all young people who earned a high school diploma
c. report the 95% confidence interval for the percentage of all young people who earned a high school diploma
2 Answers
- cidyahLv 74 years agoFavorite Answer
p^ = 0.87
a)
Standard error of estimate of the percentage of all young people who earned a high school diploma :
= sqrt ( p^(1-p^) / n)
= sqrt ( (0.87) (0.13) /1200) = 0.00971
b)
Margin of error = z * SE
z= 1.96 for 95% confidence interval
Margin of error = (1.96)(0.000971) = 0.001903
c)
Sample proportion phat = 0.87
Variance of proportion = p*(1-p)/n
= 0.87(0.13)/1200 =0.0000943
S.D. of p is sqrt[0.000094] = 0.0097
Confidence interval:
phat-zval*sd = 0.87 - (1.96)(0.009708)
phat-zval*sd = 0.87 + (1.96)(0.009708)
Confidence interval is ( 0.851 , 0.889 )
