Chapter 7 Estimation

Transcription

1 Chapter 7 Estimation

2 Point Estimate an estimate of a population parameter given by a single number

3 Examples of Point Estimates x ü is used as a point estimate for µ. ü s is used as a point estimate for σ.

4 Error of Estimate the magnitude of the difference between the point estimate and the true parameter value

5 The error of estimate using x as a point estimate for µ is: x µ

6 Confidence Level A confidence level, c, is a measure of the degree of assurance we have in our results. The value of c may be any number between zero and one. Typical values for c include 0.90, 0.95, and 0.99.

7 Critical Value for a Confidence Level, c the value z c such that the area under the standard normal curve falling between z c and z c is equal to c.

8 Critical Value for a Confidence Level, c P( z c < z < z c ) = c -z c z c

9 Find z 0.90 such that 90% of the area under the normal curve lies between z and z z 0.90 z 0.90 P(-z 0.90 < z < z 0.90 ) = 0.90

10 Find z 0.90 such that 90% of the area under the normal curve lies between z and z z 0.90 z 0.90 P(0< z < z 0.90 ) = 0.90/2 =

11 Find z 0.90 such that 90% of the area under the normal curve lies between z and z 0.90 According to Table 4 in Appendix I, lies roughly halfway between two values in the table (.4495 and.4505). Calculating the invnorm(0.05) gives you the critical value of z 0.90 =

12 Common Levels of Confidence and Their Corresponding Critical Values Level of Confidence, c Critical Value, z c 0.70 or 70% or 75% or 80% or 85% or 90% or 95% or 98% or 99% or 99.9%

13 Confidence Interval for the Mean of Large Samples (n 30) x E < µ < x + E where x = Sample Mean E = z c σn if the population standard deviations is known

14 Confidence Interval for the Mean of Large Samples (n 30) The answer is expressed in a sentence. The form of the sentence is given below. We can say with a c% confidence level that (whatever the problem is about) is between x z c sn and x + z c s n units.

15 Create a 95% confidence interval for the mean driving time between Philadelphia and Boston. Assume that the mean driving time of 64 trips was 5.2 hours with a standard deviation of 0.9 hours.

16 Key Information x = 5.2 hours s = 0.9 hours c = 95%, so z c = n = 64

17 95% Confidence Interval: < µ < < µ < < µ < < µ < We can say with 95% a confidence level that the population mean driving time from Philadelphia to Boston is between and hours.

18 95% Confidence Interval: Calculator Computation We can say with a 95% confidence level that the population mean driving time from Philadelphia to Boston is between and hours. STAT TESTS # 7 ZInterval Inpt : Stats σ = 0.9 x = 5.2 n = 64 C Level : 0.95 Calculate ( , )

19 When estimating the mean, how large a sample must be used in order to assure a given level of confidence? Use the formula: 2 z cσ n = E

20 Determine the sample size necessary to determine (with 99% confidence) the mean time it takes to drive from Philadelphia to Boston. We wish to be within 15 minutes of the true time. Assume that a preliminary sample of 45 trips had a standard deviation of 0.8 hours.

21 ... determine with 99% confidence... z 0.99 =

22 ... We wish to be within 15 minutes of the true time.... E = 15 minutes or E = 0.25 hours

23 ...a preliminary sample of 45 trips had a standard deviation of 0.8 hours. Since the preliminary sample is large enough, we can assume that the population standard deviation is approximately equal to 0.8 hours. σ = 0.8

24 Minimum Required Sample Size n = n = z cσ E 2 ( ) n = n = ( ) 2 n = n = 68 ( ) 2 2

25 Rounding Sample Size Any fractional value of n is always rounded to the next higher whole number.

26 We would need a sample of 68 trip times to have a 99% confidence level for the population mean time it takes to drive from Philadelphia to Boston with an error of 0.25 hours.

27 THE END OF THE PRESENTATION

28 Answers to the Sample Questions

29 1. As part of a study on AP test results, a local guidance counselor gathered data on 200 tests given at local high schools. The test results are based on scores of 1 to 5, where a 1 means a very poor test result to a 5 which means a superior test result. The sample mean was 3.62 with a standard deviation of a. Construct a 90% confidence interval for the population mean.

30 ( ) x z c σn < µ < x + z c σ n < µ < ( ) < µ < < µ < < µ < We can say with a 90% confidence level that the population mean score on AP tests at local high schools is between and

31 1. As part of a study on AP test results, a local guidance counselor gathered data on 200 tests given at local high schools. The test results are based on scores of 1 to 5, where a 1 means a very poor test result to a 5 which means a superior test result. The sample mean was 3.62 with a standard deviation of b. Construct a 95% confidence interval for the population mean.

32 ( ) x z c σn < µ < x + z c σ n < µ < ( ) < µ < < µ < < µ < We can say with a 95% confidence level that the population mean score on AP tests at local high schools is between and

33 1. As part of a study on AP test results, a local guidance counselor gathered data on 200 tests given at local high schools. The test results are based on scores of 1 to 5, where a 1 means a very poor test result to a 5 which means a superior test result. The sample mean was 3.62 with a standard deviation of c. Perform the calculator checks for parts a and b.

34 1. Part c - Calculator check for part a ZINTERVAL Input :Stats σ :0.84 x : 3.62 n :200 C Level :0.90 Calculate ZINTERVAL ( , ) x = 3.62 n = 200 VARS STATISTICS TEST H :lower = I :upper =

35 1. Part c - Calculator check for part b ZINTERVAL Input :Stats σ :0.84 x : 3.62 n :200 C Level :0.95 Calculate ZINTERVAL ( , ) x = 3.62 n = 200 VARS STATISTICS TEST H :lower = I :upper =

36 1. As part of a study on AP test results, a local guidance counselor gathered data on 200 tests given at local high schools. The test results are based on scores of 1 to 5, where a 1 means a very poor test result to a 5 which means a superior test result. The sample mean was 3.62 with a standard deviation of d. How many test results would be require to be 95% confident that the sample mean test score is within 0.05 of the population mean score?

37 n = n = z cσ E 2 ( ) ( 0.84) n = n = ( ) 2 n = n = We would need to acquire 1,085 AP test results to have a 95% confidence level with an error of no more than 0.05 for the population mean AP test scores.

38 2. The SAT results for 50 randomly selected seniors are listed below. The score are based only on the English and Math portions of the SAT examination. Use your calculator to determine a 99% confidence interval for the mean score of the SAT examination

39 Remember that you need perform a 1-VAR- STATS calculation on the data first to get the value for the sample standard deviation. ZINTERVAL Input : Data σ : List : L 1 Freq :1 C Level :0.99 Calculate ZINTERVAL ( , ) x = n = 50 VARS STATISTICS TEST H :lower = I :upper =

40 We can say with a 99% confidence level that the population mean score on the SAT examination of the seniors at a local high school is between and

41 THE END OF SECTION 1