IS) Marina reads that 89% of U.S. school kids .-- attend public schools, 8% attend private schools, and 3% are homeschooled. She attains school attendance information for a sample of 170 school-aged...

The quality control manager at a light bulb factory needs to estimate the mean life of a large
shipment of light bulbs. The population standard deviation is known to be 100 hours. A random
sample of 64 light bulbs indicated a sample mean life of 350 hours.
1. We wish to determine the best point estimate of the population mean life of the light
bulbs in the shipment.
The best estimate for the population mean is the sample mean, i.e., 350 hours.
2. In order to set up a confidence interval for the population mean life of the light bulbs in
this shipment, we should use z, since t is typically used for sample sizes of 30 or less. For
larger sample sizes, z gives a good estimate of the confidence interval.
3. We wish to set up a 90% confidence interval estimate for the population mean life of the
light bulbs in the shipment.
A 90% confidence interval includes all light bulbs whose life is between 5 and 95
percentile. Thus we first must determine the standard score (z-value) of a light bulb
whose life is 95 percentile. From a standard normal distribution table [1], we find that
this z-value is 1.65. Thus, the 90% confidence interval goes from 1.65 standard
deviations below the mean life to 1.65 standard deviation above the mean life, i.e., from
350 – 165 = 185 hours to 350 + 165 = 515 hours. Thus, the 90% confidence interval is
[185 hours, 515 hours].
4. We wish to set up a 95% confidence interval estimate for the population mean life of the
light bulbs in the shipment.
A 95% confidence interval includes all light bulbs whose life is between 2.75 and 97.5
percentile. Thus we first must determine the standard score (z-value) of a light bulb...