Exam 9: Statistical Inferences Based on Two Samples

Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS)of their patients.In 1996,the average LOS for non-heart patient was 4.6 days.A random sample of 20 hospitals had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days.Assume LOS is normally distributed.A 95% confidence interval for the population mean LOS for non-heart patients in hospitals for 2000 would be _____.

A company is interested in estimating $\mu$ , the mean number of days of sick leave taken by its employees. The firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days. From historical records, the population standard deviation is about 10 days. -A 93% confidence interval for $\mu$ ,the mean number of days of sick leave would be _____.

A statistical quality control process for cereal production measures the weight of a cereal box.The population standard deviation is known to be .05 grams.In order to achieve a 97% confidence with a margin of error of .02g,how large a sample should be used?

The quality control manager at a cell phone battery factory is interested in the mean life of a large shipment of batteries.Battery life is normally distributed with a standard deviation of 100 hours.A random sample of 64 batteries indicates a sample mean of 350 hours.The 99% confidence interval for mean battery life is _____.

The provincial transportation department is studying traffic patterns on one of the busiest highways in the province.As part of the study,the department needs to estimate the average number of vehicles that pass an intersection each day to within 300 cars with 99% confidence.If the population standard deviation is 1,010 cars,how many days do they need to sample?

A random sample of size 30 from a normal population yields $\bar { x }$ = 32.8 and s = 4.51.Construct a 95 percent confidence interval for $\mu$ .

A random sample of size 10 is taken from a population assumed to be normal,and $\bar { x }$ = 1.2 and s = .6.Calculate a 95 percent confidence interval for $\mu$ .

You are studying two normally distributed populations with equal variances.A random sample of size 10 is taken from each population.The sample from the first population gives the following measurements: 16,14,19,18,19,20,15,18,17,18.The sample from the second population gives the following measurements: 13,19,14,17,21,14,15,10,13,15.Compute the 95% confidence interval for the difference between two population means

When the sample size and the sample proportion $\hat { p }$ Remain the same,a 90% confidence interval for a population proportion p will be ____________ the 99% confidence interval for p.

An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil.Three contaminated soil samples were treated.After 95 days,the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1.5%.If the percentage of contamination is normally distributed,the mean percentage of contamination removed at a 98% confidence level would be _____.

Assuming that the difference is normally distributed,find a 98 percent confidence interval for the paired difference where $\bar { d }$ = 1.6, $s _ { d } ^ { 2 }$ = 40.96,and n = 30.

A random sample of size 10 is taken from a population assumed to be normal,and $\bar { x }$ = 1.2 and the variance is .36.Calculate a 99 percent confidence interval for $\mu$ .

The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue).For this quarter,the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent.What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company?

In a manufacturing process,we are interested in measuring the average length of a certain type of bolt.Based on a preliminary sample of 9 bolts,the sample standard deviation is 0.3cm.If the lengths are normally distributed,how many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within 0.02cm of the true mean bolt length?

Let p₁ represent the population proportion of Canadian women who are in favour of a new modest tax on "junk food".Let p₂ represent the population proportion of Canadian men who are in favour of a new modest tax on "junk food".Out of the 265 women surveyed,106 of them are in favour of a "junk food" tax.Out of the 285 men surveyed,only 57 of them are in favour a "junk food" tax. -Find a 95 percent confidence interval for the difference between p₁ and p₂.

Consider a normally distribution population with known variance.A 95% confidence interval is computed based on sample of 25 observations.If we take a new sample of 100 observations and compute a second 95% confidence interval based on the new sample,we know that the width of the second confidence interval will be exactly ____ times the width of the first confidence interval.

A sample of 100 items has a standard deviation of 5.1 and a mean of 21.6.Construct a 95 percent confidence interval for $\mu$ .

A random sample of 1000 customers at a fast food restaurant had their service time recorded.Based on this sample,a 95% confidence interval for the average service time,in minutes,at the restaurant is computed to be [4.1,7.5].How would you interpret this interval?

The mid-distance running coach,Zdravko Popovich,for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run.Five mid-distance runners were randomly selected before they were trained with coach Popovich's six-month training program and their completion time of 1500-meter run was recorded (in minutes).After six months of training under coach Popovich,the same five runners' 1500 meter run time was recorded again the results are given below.Assuming the difference in running times is normally distributed. Runner 1 2 3 4 Completion time before training 5.9 7.5 6.1 6.8 Completion time after training 5.4 7.1 6.2 6.5 -Construct the appropriate 95% confidence interval for the average change in completion time.

A sample of 2,000 people yielded $\hat { p }$ =.52. -What is the estimated standard deviation of the sample proportion?