Exam 9: Statistical Inferences Based on Two Samples

Find a 99 percent confidence interval for p when $\hat { p }$ = .51 and n = 1,000.

In a survey of 1,000 people,420 are opposed to the tax increase.Construct a 95 percent confidence interval for the proportion of those people opposed to the tax increase.

A 95% confidence interval for a population proportion is computed to be [0.14,0.36].This means that each time we take a sample from this population,the sample proportion has a 95% chance of being between 0.14 and 0.36.

The customer service manager for the XYZ Fastener Manufacturing Company examined 60 vouchers and found nine vouchers containing errors.Find a 95 percent confidence interval for the proportion of vouchers with errors.

As the stated confidence level decreases,the width of the confidence interval __________.

A Research and Development Laboratory researcher for a paint company is measuring the level of a certain chemical contained in a certain type of paint.If the paint contains too much of this chemical,the quality of the paint will be compromised.How many cans of paint should the sample contain if the researcher wants to be 95% certain of being within 1% of the true proportion of this chemical?

Find a 99 percent confidence interval for $\mu$ if $\bar { x }$ = 98.6,s = 2,and n = 5.Assume that the sample is randomly selected from a normally distributed population.

Find the 99 percent confidence interval for p when $\hat { p }$ = 0.2,and n = 100.

Consider a confidence interval for a population mean.As the standard deviation ________,sample sizes need to _____________ to achieve a specified margin of error.

The success rate of a procedure is 37 per 120 cases in a sample.Find a 95 percent confidence interval for the actual success proportion of the procedure.

In the construction of a confidence interval,as the confidence level required in estimating the mean increases,the width of the confidence interval for a population mean ______________.

At the end of 1990,1991,and 1992 the average prices of a share of stock in a money market portfolio were $34.83,$34.65 and $31.26 respectively.To investigate the average share price at the end of 1993,a random sample of 30 stocks was drawn and their closing prices on the last trading day of 1993 were observed with a mean of 33.583 and a standard deviation of 19.149.Assuming that stock prices are normally distributed,a 90% confidence interval would be _____.

Find a 95 percent confidence interval for p when $\hat { p }$ =.25 and n = 400.

A bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years.A random sample of 20 defaulted loans yielded a mean of $67,918 with a standard deviation of $16,552.40.If the outstanding balance is normally distributed,the 90% confidence interval for the mean balance of defaulted loans over the past three years would be _____.

A market research company is interested in the percentage of all customers to a department store that visit the shoe department.They conduct a survey and finds that 79 out of 384 customers visit the shoe department.Calculate a 98% confidence interval for the true proportion all customers to a department store that visit the shoe department.

Unoccupied seats on flights cause airlines to lose revenue.Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year.A preliminary study of 10 flights reveals a sample standard deviation of 6.3 empty seats.How many flights should we select if we wish to estimate $\mu$ To within 1 seats and be 95% confident? Assume the number of empty seats can be approximated by a normal distribution.

An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company.A random sample of 240 claims shows that the insurance company covered 90 accident claims while 150 claims were not covered.Use a confidence interval of 95% and determine the margin of error for the estimate.

We wish to compute a confidence interval for a population mean.Suppose we are determining the sample size n needed to achieve a desired margin of error for this interval.If we compute the value of n to be 79.2,we would choose to sample 79 observations.

The quality control manager of a tire company wishes to estimate the tensile strength of a standard size of rubber used to make a class of radial tires.A random sample of 61 pieces of rubber from different production batches is subjected to a stress test.The test measures the force needed to break the rubber.According to the sample results,the average pressure is 238.4 and the standard deviation is 35.Determine the 98% confidence interval.

A sample of 2,000 people yielded $\hat { p }$ =.52. -Calculate a 90 percent confidence interval for p.