1. The quality control manager at a lightbulb factory needs to estimate the mean
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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
Mean is 450 and standard deviation s is 60. thus standard error is SE=s/sqrt(N)=60/sqrt(30)=10.95
z for 98% confidence is 2.33
thus lower limit is mean-z*SE=450-2.33*10=426.7
upper limit is mean+s*SE=450+2.33*10=473.3
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