Question

3. In a recent study of 100 eighth graders, the mean number of hours per week that they watched television was 22.6. Assume the population standard deviation, o, is 5.2 hours. (Section 6.1) Construct a 90% confidence interval for the population mean number of hours that eighth graders watch television • Margin of error, E..8554 Confidence interval:al: 74<us 23.46 If the population standard deviation was reduced by 50% to 2.6 and the level of confidence remained at 90%, what would be the new margin of error and confidence interval? O Margin of error, E. Confidence interval: - < < O Did the confidence interval increase or decrease and why? If the population standard deviation was doubled to 10.4 and the level of confidence remained at 90%, what would be the new margin of error and confidence interval? o Margin of error, E. Confidence interval: <u< O Did the confidence interval increase or decrease and why?

Answer 1

a)

sample mean sample size population std deviation 22.6000 100.00 5.200 standard error of mean = Ox=0/Vn= 0.5200 ne for 90 % CI value of z= margin of error E=z*std error lower confidence bound=sample mean-margin of error= Upper confidence bound=sample mean +margin of errors 1.645 0.86 21.745 23.455

margin of error =0.86 , confidence interval: 21.74 <μ <23.46

b)

margin of error =0.43 , confidence interval: 21.17 <μ <23.03

Confidence interval decrease as spread of the distribution decreases

c)

margin of error =1.71 , confidence interval: 20.89 <μ <24.31

Confidence interval increase as spread of the distribution increases