McGraw-Hill/IrwinCopyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved. Confidence Intervals Chapter 8

8-2 Confidence Interval A confidence interval for the population mean is an interval constructed around the sample mean so that the interval will contain the population mean with certain probability. The probability that the interval contains the population parameter is called a confidence level. The probability for the interval not to contain the population mean is 1- confidence level, denoted by .  confidence level confidence level=1- 

8-3 Example 8.1: The Car Mileage Case Suppose an automaker conducts mileage tests on a sample of 50 of its new mid- size cars and obtains the sample mean with  = Assuming population standard deviation σ=0.8. Please compute the 95 percent confidence interval of the population mean.

8-4 Confidence Interval vs  The probability that the confidence interval will contain the population mean, the confidence level, is denoted by  If  what is the confidence level?   If the confidence level is 99%,  ?

8-5 z-Based Confidence Intervals for a Population Mean: Population σ is Known –If a population is normally distributed with mean  and standard deviation σ, then the sampling distribution of  is normal with mean   =  and standard deviation –  is approximately normally distributed if population mean  standard deviation σ both exist and n≥30.

8-6 z-Based Confidence Intervals for a Population Mean: Population σ is Known If a population has standard deviation  (known), and if the population is normal or if sample size is large (n  30), then … … a (1-  ) confidence interval for  is

8-7 t and Right Hand Tail Areas The definition of the critical value Z α ZαZαZαZα The area to the right if 1-α Z α is the percentile such that P(Z< Z α ) =1-α

8-8 Find out the critical values Z α/2 Confidence level α α/2 P( Z < Z α/2 ) =1- α/2 P(Z<Z α/2 ) = P(Z<Z α/2 ) = P(Z<Z α/2 ) = P(Z<Z α/2 ) = P(Z<Z α/2 ) = Z α/2 Z 0.05 = Z = Z =Z 0.02 =Z 0.03 =

8-9 Find out the critical values Z α/2 Confidence level α α/ P( Z < Z α/2 ) =1- α/2 P(Z<Z α/2 ) =0.95 P(Z<Z α/2 ) =0.975 P(Z<Z α/2 ) =0.995 P(Z<Z α/2 ) =0.98 P(Z<Z α/2 ) =0.97 Z α/2 Z 0.05 = Z = 1.96 Z = Z 0.02 = 2.05 Z 0.03 = 1.88

 Confidence Interval If the population is normal or if n≥30, and population standard deviation is know, the 1-  confidence interval for population mean is –The normal point z  /2 gives a right hand tail area under the standard normal curve equal to  /2

8-11 Example 8.1: The Car Mileage Case Suppose an automaker conducts mileage tests on a sample of 50 of its new mid- size cars and obtains the sample mean with  = Assuming population standard deviation σ=0.8. Please compute the 95 percent confidence interval of the population mean.  = 31.56;σ = 0.8;n = 50

% Confidence Interval A bank manager developed a new system to reduce the service time. Suppose the new service time has a normal distribution with known standard deviation 2.47 minutes. The mean of a sample of 10 randomly selected customers is Please construct an 99% confidence interval of the population mean.

% Confidence Interval For a 99% confidence level, 1 –  = 0.99, so  = 0.01, and  /2 = –Reading between table entries z = The 99% confidence interval is

% Confidence Interval  = 5.46;σ = 2.47;n = 10

8-15 Notes on the Example The confidence interval can be expression as margin of error The length of confidence interval is equal to 2E. The 99% confidence interval is slightly wider than the 95% confidence interval –The higher the confidence level, the bigger the critical value Z α/2, the wider the interval.

8-16 The Effect of a on Confidence Interval Width z  /2 = z = 1.96z  /2 = z = 2.575

8-17 t-Based Confidence Intervals for a Mean:  Unknown If  is unknown (which is usually the case), we can construct a confidence interval for  based on the sampling distribution of If the population is normal, then for any sample size n, this sampling distribution is called the t distribution

8-18 t-Based Confidence Intervals for a Mean:  Unknown If the sampled population is normally distributed with mean , then a (1­  )100% confidence interval for  is The result applies if sample size is ≥30 t  /2 is the t point giving a right-hand tail area of  /2 under the t curve having n­1 degrees of freedom

8-19 The t Distribution The curve of the t distribution is similar to that of the standard normal curve –Symmetrical and bell-shaped –The t distribution is more spread out than the standard normal distribution –The spread of the t is given by the number of degrees of freedom Denoted by df For a sample of size n, there are one fewer degrees of freedom, that is, df = n – 1

8-20 Degrees of Freedom and the t-Distribution As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve

8-21 The t Distribution and Degrees of Freedom As the sample size n increases, the degrees of freedom also increases As the degrees of freedom increase, the spread of the t curve decreases As the degrees of freedom increases indefinitely, the t curve approaches the standard normal curve –If n ≥ 30, so df = n – 1 ≥ 29, the t curve is very similar to the standard normal curve

8-22 t and Right Hand Tail Areas Use a t point denoted by t  –t  is the point on the horizontal axis under the t curve that gives a right hand tail equal to a –So the value of t  in a particular situation depends on the right hand tail area a and the number of degrees of freedom df = n – 1  = 1 – a, where 1 – a is the specified confidence coefficient

8-23 Definition of the critical value t α

8-24 Using the t Distribution Example: Find t  for a sample of size n=15 and right hand tail area of –For n = 15, df = 14, t =2.145 –  = Note that a = corresponds to a confidence level of 0.95 –In Table 8.3, along row labeled 14 and under column labeled 0.025, read a table entry of –So t  = 2.145

8-25 Using the t Distribution Continued

8-26 Find out the critical values t α/2 Confidence level and degree of freedom Df=10, CL=0.9 Df=20, CL=0.95 Df=25, CL=0.99 Df=30, CL=0.98 Df=40, CL=0.99 α α/ P(t>t α/2 ) = α/2 P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = t α/2 t 0.05 =1.812

8-27 Find out the critical values Z α/2 Confidence level and degree of freedom Df=10, CL=0.9 Df=20, CL=0.95 Df=25, CL=0.99 Df=30, CL=0.98 Df=40, CL=0.99 α α/2 P(t<t α/2 ) =? P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = P(t<t α/2 ) = t α/2 t 0.05 = t = t = t 0.01 = t = 2.704

8-28 Suppose that, in order to reduce risk, a large bank has decided to initiate a policy limiting the mean debt-to-equity ratio for its portfolio of commercial loans to 1.5. In order to estimate the mean debt-to- equity ratio of its load portfolio, the bank randomly selects a sample of 15 of its commercial loan accounts. The sample mean and standard deviation of the sample is and The population has normal distribution. Please construct a 95% confidence interval for the mean of the debt-to-equity ratio.

8-29 Example 8.4 Debt-to-Equity Ratios Estimate the mean debt-to-equity ratio of the loan portfolio of a bank Select a random sample of 15 commercial loan accounts –  = – s = – n = 15 Want a 95% confidence interval for the ratio Assume all ratios are normally distributed but σ unknown

8-30 Example 8.4 Debt-to-Equity Ratios Continued Have to use the t distribution At 95% confidence, 1 –  = 0.95 so  = 0.05 and  /2 = For n = 15, df = 15 – 1 = 14 Use the t table to find t  /2 for df = 14, t  /2 = t = The 95% confidence interval:

8-31 Length of Confidence Interval The confidence interval can be expression as margin of error equal to or The length of confidence interval is equal to 2E.

8-32 Sample Size Determination (z) If σ is known, then a sample of size of at least will result in a confidence interval such so that  is within E units of  with probability 100(1-  )%.

8-33 Sample Size Determination (t) If σ is unknown and is estimated from s, then a sample of size of at least will give an interval so that  is within E units of , with 100(1-  )% confidence. The number of degrees of freedom for the t  /2 point is the size of the preliminary sample minus 1.

8-34 A bank manager developed a new system to reduce the service time. Suppose the new service time has a normal distribution with known standard deviation 2.47 minutes. How large the sample should be if the manager wants to be 99% confident that sample mean is within 0.5 minute of mu (the population mean). round up to 162

8-35 A new alert system is installed for air traffic controllers. It is hoped that the mean “alert time” for the new equipment is less than 8 seconds. In order to test the equipment 15 randomly selected air traffic controllers are trained to use the machine and their alert times for a simulated collision course are recorded. The sample alert times has a mean of 7.4 seconds and s= Supposed the alert times are normally distributed, please construct 95% confidence interval for the mean alert time of the machine. Can we be 95% confident that mu is less than 8 seconds? Determine the sample size needed to make us 95% confident that the sample mean is within a margin of error of 0.3 second of mu. t =2.145, for degree of freedom 14; rockville

8-36 Selecting an Appropriate Confidence Interval for a Population Mean