1) Find the margin of error in the 95% confidence interval for p that is calcula

Question

1) Find the margin of error in the 95% confidence interval for p that is calculated from the sample statistics x = 207 and n = 300.

2) Use technology to find the critical value t/2 that corresponds to a confidence level of 80% from a sample of size n = 32.

3) Find the margin of error in the 95% confidence interval for that is calculated from a sample with mean of 18.5, standard deviation of 2.0, and size of 9. Assume that the population is approximately normal and that is unknown.

4) Find the margin of error in the 95% confidence interval for that is calculated from a sample with mean of 5.86, standard deviation of 0.27, and size of 25. Assume that the population is normally distributed and is known to be 0.23

I'm having alot of trouble with these. Thank you so much in advance!

Explanation / Answer

1)here p=x/n=0.69

therefore std error =(p(1-p)/n)1/2 =0.0267

for 95% CI; z=1.96

therefore margin of error =z*std error =0.0523

2) from t value table: crtiical value = -/+1.3095

3)

std error =std deviation/(n)1/2 =2/(9)1/2 =0.6667

for 95% CI and (n-1=8) degree of freedom ; t=2.306

hence margin of error =t*std error =1.5373

4)

std error =std deviation/(n)1/2 =0.23/(25)1/2 =0.046

for 95% CI ; z=1.96

hence margin of error =z*std error =0.0902

5)

std error =std deviation/(n)1/2 =2/(45)1/2 =0.2981

for 90% CI and (n-1=44) degree of freedom ; t=1.6802

hence margin of error =t*std error =0.5009

hence 90% confidence interval =sample mean -/+ margin of error =3.6991 ; 4.7009

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