1. The quality control manager at a lightbulb factory needs to estimate the mean
ID: 3340654 • Letter: 1
Question
1. The quality control manager at a lightbulb factory needs to estimate the mean life of a new type of lightbulb. The population standard deviation is assumed to be 60 hours. A random sample of 30 lightbulbs shows a sample mean life of 450 hours. Construct and explain a 98% confidence interval estimate of the population mean life of the new lightbulb.
2. A survey of first-time home buyers found that the sample mean annual income was $48,000. Assume that the survey used a sample of 25 first-time home buyers and that the sample standard deviation was $1,150. Compute and explain a 95% confidence interval estimate of the population mean.
3. A telephone poll of 800 American adults asked "where would you rather go in your spare time?" One response, by 275 adults, was "a movie". Compute and explain a 90% confidence interval estimate of the proportion of all American adults who would respond "a movie".
Explanation / Answer
(1) (1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)
98% confidence interval for population mean=450±z(0.02/2)*60/sqrt(30)=450±2.32*60/sqrt(30)=
=450±25.4=(424.6, 475.4)
(2)(1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*sd/sqrt(n)
95% confidence interval for population mean=48000±t(0.05/2, 25-1)*1150/sqrt(25)=
=48000±2.06*1150/sqrt(25)=48000±474=(47526,48474)
(3) p=x/n=275/800=0.3438
SE(p)=sqrt(p(1-p)/n)=sqrt(0.3438*(1-0.3438)/800)=0.0168
(1-alpha)*100% confidence interval for population proportion (P)=sample proportion (p) ±z(alpha/2)*SE(p)
90% confidence interval =0.3438±z(0.1/2)*0.0168=0.3438±1.645*0.0168=0.3438±0.0276=(0.3162,0.3714)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.