1. The quality control manager at a lightbulb factory needs to estimate the mean

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 80 hours. A random sample of 40 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 28 first-time home buyers and that the sample standard deviation was $2,150. Compute and explain a 95% confidence interval estimate of the population mean.

3. A telephone poll of 900 American adults asked "where would you rather go in your spare time?" One response, by 250 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) here std error =std deviation/(n)1/2 =12.649

for 98% CI, z=2.3263

hence confidence interval =mean -/+ z*std errror =420.5738 ; 479.4262

2)here std error =std deviation/(n)1/2 =406.3118

for 95% CI, t=2.0518

hence confidence interval =mean -/+ t*std errror =47166.32 ; 48333.68

3)here p=250/900=0.2778

std error =(p(1-p)/n)1/2 =0.0149

for 90% CI, z=1.6449

hence confidence interval =p -/+ z*std error =0.2532 ; 0.3023

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