1. The quality control manager at a light bulb factory needs to estimate the mean life of a new type of light bulb. The population standard deviation is assumed to be 60 hours. A random sample of 32 lights shows a sample mean life of 490 hours. Construct and explain a 95% confidence inverval estimate of the population mean life of the new light bulb.

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1. The quality control manager at a light bulb factory needs to estimate the mean life of a new
type of light bulb. The population standard deviation is assumed to be 60 hours. A random
sample of 32 lights shows a sample mean life of 490 hours. Construct and explain a 95%
confidence inverval estimate of the population mean life of the new light bulb.
2. A survey of first time home buyers found that the sample mean annual income was $46,300.
Assume that the survey used a sample of 28 first time home buyers and that the sample
standard deviation was $1,100. Compute and explain a 90% confidence interval estimate of
the population mean.
3. A telphone poll of 850 American adults asked "where would you rather go i your spare time?
One response by 255 adults, was "the mall'. Compute and explain a 98% confidence interval
estimate of the proportion of all American adults would respond "the mall".
4. For purpose #1l above size sample would be needed to achieve a margin of error of 15 hours
or less?
Transcribed Image Text:1. The quality control manager at a light bulb factory needs to estimate the mean life of a new type of light bulb. The population standard deviation is assumed to be 60 hours. A random sample of 32 lights shows a sample mean life of 490 hours. Construct and explain a 95% confidence inverval estimate of the population mean life of the new light bulb. 2. A survey of first time home buyers found that the sample mean annual income was $46,300. Assume that the survey used a sample of 28 first time home buyers and that the sample standard deviation was $1,100. Compute and explain a 90% confidence interval estimate of the population mean. 3. A telphone poll of 850 American adults asked "where would you rather go i your spare time? One response by 255 adults, was "the mall'. Compute and explain a 98% confidence interval estimate of the proportion of all American adults would respond "the mall". 4. For purpose #1l above size sample would be needed to achieve a margin of error of 15 hours or less?
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