1. 1 ES Chapter 18 & 20: Inferences Involving One Population Student’s t, df = 5 Student’s t, df = 15 Student’s t, df = 25

2. 2 ES Chapter Goals Learned about confidence intervals Assumed  was known Consider inference about  when  is unknown Consider inference about p, the probability of success

3. 3 ES Inferences about  are based on the sample mean If the sample size is large or the sample population is normal: has a standard normal distribution )//()(*nxz   Inference About mean  (  unknown) If  is unknown, use s as a point estimate for  Estimated standard error of the mean:

4. 4 ES 1.When s is used as an estimate for , the test statistic has two sources of variation: sx and 4.The population standard deviation, , is almost never known in real-world problems The standard error will almost always be estimated using Almost all real-world inference about the population mean will be completed using the Student’s t-statistic Student’s t-Statistic 3.Assumption: samples are taken from normal populations 2.The resulting test statistic: Known as the Student’s t-statistic

5. 5 ES Properties of the t-Distribution (df>2) 3.t is distributed so as to form a family of distributions, a separate distribution for each different number of degrees of freedom 1.t is distributed with a mean of 0 2.t is distributed symmetrically about its mean 4.The t-distribution approaches the normal distribution as the number of degrees of freedom increases 5.t is distributed with a variance greater than 1, but as the degrees of freedom increase, the variance approaches 1 6.t is distributed so as to be less peaked at the mean and thicker at the tails than the normal distribution

6. 6 ES Student’s t, df = 5 Student’s t, df = 15 Normal distribution Degrees of Freedom, df: A parameter that identifies each different distribution of Student’s t-distribution. For the methods presented in this chapter, the value of df will be the sample size minus 1, df = n  1. Student’s t-Distributions

7. 7 ES 1.The number of degrees of freedom associated with s 2 is the divisor (n  1) used to define the sample variance s 2 Thus: df = n  1 Notes 2.The number of degrees of freedom is the number of unrelated deviations available for use in estimating  2 3.Table for Student’s t-distribution (Table C) is a table of critical values. Left column = df. When df > 100, critical values of the t- distribution are the same as the corresponding critical values of the standard normal distribution. 4.Notation: t (df,  ) Read as: t of df, 

8. 8 ES ) df,  ( t t-Distribution Showing t (df,  )

9. 9 ES Example Example:Find the value of t (12, 0.025) Portion of Table 6......... - t (12, 0.025) t (12, 0.025)

10. 10 ES 1.If the df is not listed in the left-hand column of Table C, use the next smaller value of df that is listed Notes cumulative probability 2.Most computer software packages will calculate either the area related to a specified t-value or the t-value that bounds a specified area 3.The cumulative distribution function (CDF) is often used to find area from to t 4.If the area from to t is known and the value of t is wanted, then the inverse cumulative distribution function (INVCDF) is used

11. 11 ES Confidence Interval Procedure: 1.Procedure for constructing confidence intervals similar to that used when  is known The assumption for inferences about mean  when  is unknown: The sampled population is normally distributed The Assumption... 2.Use t in place of z, use s in place of  3.The formula for the  confidence interval for  is: 1df wheret (df,  to  n n s x n s xt (df, 

12. 12 ES Example:A study is conducted to learn how long it takes the typical tax payer to complete their federal income tax return. A random sample of 17 income tax filers showed a mean time (in hours) of 7.8 and a standard deviation of 2.3. Find a 95% confidence interval for the true mean time required to complete a federal income tax return. Assume the time to complete the return is normally distributed. Solution: 1.Parameter of Interest The mean time required to complete a federal income tax return Example 2.Confidence Interval Criteria a.Assumptions: Sampled population assumed normal,  unknown b.Test statistic: t will be used c.Confidence level:  = 0.95

13. 13 ES 3.The Sample Evidence: Solution Continued 5.The Results: 6.62 to 8.98 is the 95% confidence interval for  4.The Confidence Interval a.Confidence coefficients: b.Maximum error: c.Confidence limits: 18.1)5578.0)(12.2( 17 3.2 )12.2(  n s E t (df,  /2) = t (16, 0.025) = 2.12 t (16, 0.025)

14. 14 ES Inferences About the Probability of Success Possibly the most common inference of all Many examples of situations in which we are concerned about something either happening or not happening Two possible outcomes, and multiple independent trials

15. 15 ES 1.p: the binomial parameter, the probability of success on a single trial Background 2. : the observed or sample binomial probability 'p x represents the number of successes that occur in a sample consisting of n trials 3.For the binomial random variable x: 4.The distribution of x is approximately normal if n is larger than 20 and if np and nq are both larger than 5

16. 16 ES 1.a mean equal to p, 2.a standard error equal to, and 3.an approximately normal distribution if n is sufficiently large 'p  'p  Sampling Distribution of p': If a sample of size n is randomly selected from a large population with p = P(success), then the sampling distribution of p' has: Sampling Distribution of p' In practice, use of the following guidelines will ensure normality: 1.The sample size is greater than 20 2.The sample consists of less than 10% of the population 3.The products np and nq are both larger than 5

17. 17 ES The assumptions for inferences about the binomial parameter p: The n random observations forming the sample are selected independently from a population that is not changing during the sampling The Assumptions... Confidence Interval Procedure: The unbiased sample statistic p' is used to estimate the population proportion p The formula for the  confidence interval for p is: n qp p n qp z (  p '' ' to '' '  z ( 

18. 18 ES Example: A recent survey of 300 randomly selected fourth graders showed 210 participate in at least one organized sport during one calendar year. Find a 95% confidence interval for the proportion of fourth graders who participate in an organized sport during the year. Solution: 1.Describe the population parameter of concern The parameter of interest is the proportion of fourth graders who participate in an organized sport during the year Example 2.Specify the confidence interval criteria a.Check the assumptions The sample was randomly selected Each subject’s response was independent

19. 19 ES b.Identify the probability distribution z is the test statistic p' is approximately normal c.Determine the level of confidence:  0.95 Solution Continued 3.Collect and present sample evidence Sample information: n = 300, and x = 210 The point estimate:

20. 20 ES 4.Determine the confidence interval a.Determine the confidence coefficients: Using Table A: z (  /2) = z (0.025) = 1.96 Solution Continued c.Find the lower and upper confidence limits: 0519.0)0265.0)(96.1(0007.0)96.1( 300 )30.0)(70.0( 96.1 ''   n qp E b.The maximum error of estimate: z (  /2)

21. 21 ES d.The Results 0.6481 to 0.7519 is a 95% confidence interval for the true proportion of fourth graders who participate in an organized sport during the year Solution Continued E: maximum error of estimate  : confidence level p*: provisional value of p (q* = 1  p*) If no provisional values for p and q are given use p* = q* = 0.5 (Always round up) 2 2 **][ E qp n   Sample Size Determination: z (  /2)

22. 22 ES Example: Determine the sample size necessary to estimate the true proportion of laboratory mice with a certain genetic defect. We would like the estimate to be within 0.015 with 95% confidence. Example Solution: 1.Level of confidence:  = 0.95, z (  /2) = z (0.025) = 1.96 2.Desired maximum error is E = 0.015. 3.No estimate of p given, use p* = q* = 0.5 4.Use the formula for n: 4269 44.4268 000225.0 9604.0 )015.0( )5.0()5.0()96.1(**][ 2 2 2 2      n E qp n z (  /2)

23. 23 ES 211 210.75 000225.0 0474.0 )015.0( )9875.0()0125.0()96.1(**][ 2 2 2 2      n E qp n Note:Suppose we know the genetic defect occurs in approximately 1 of 80 animals Use p* = 1/80 = 0.125: Note As illustrated here, it is an advantage to have some indication of the value expected for p, especially as p becomes increasingly further from 0.5 z (  /2)