TABLE 4.1 POPULATION OF 100 VALUES 2

Transcription

1 TABLE 4. POPULATION OF 00 VALUES WITH µ = 6. AND = PLATE 4-

2 CLASS EXPERIMENT: Randomly select data sets of 0 values and compute mean and variance for the sample. Set : Set : Set 3: Set 4: Set 5: Set 6: Set 7: Set 8: Set 9: Set 0: Solution: Set : Set : Set 3: Set 4: Set 5: Set 6: Set 7: Set 8: Set 9: Set 0: y = 5.89 y = 6.4 y = 6.03 y = 3.84 y = 4.45 y = 6.64 y = 5.96 y = 4.0 y = 5.66 y = 6.49 S = 8.4 S = 36.9 S = 9.39 S =.05 S = 0.35 S = 7.7 S = 8.65 S = 9.5 S = 8.37 S = 5.9 PLATE 4-

3 CLASS EXPERIMENT: Randomly select data sets of 0 values and compute mean and variance for the sample. Set : Set : Set 3: Set 4: Set 5: Set 6: Set 7: Set 8: Set 9: Set 0: Solution: Set : y = 5.30 S = 4.37 Set 6: y = 3.86 S =.90 Set : y = 5.48 S = 0.55 Set 7: y = 3.78 S = 5.6 Set 3: y = 5.50 S =.7 Set 8: y = 5.57 S = 8.56 Set 4: y = 5.60 S = 37.3 Set 9: y = 3.95 S = 9.40 Set 5: y = 4.53 S = Set 0: y = 6.67 S = 8.0 PLATE 4-3

4 CHANGES IN ESTIMATES FOR POPULATION MEAN AND VARIANCE WITH INCREASING SAMPLE SIZE # S y PLATE 4-4

5 CHI SQUARE DISTRIBUTION: v S,v. Distribution of sample variances. Used to build confidence intervals for population variance. 3. Starts at 0 and goes to 4. Distribution is NOT symmetric 5. Based on degrees of freedom, v, from sample., v 6. Table gives critical value which delineates area in upper tail. PLATE 4-5

6 STUDENT or t DISTRIBUTION: t y µ / n t. Distribution of sample means.. Used to build confidence intervals for population mean when (n < 30). 3. Starts at and goes to Like normal distribution, t distribution is symmetric. 5. Based on degrees of freedom, v, from sample. 6. Table gives critical t value which delineates area in upper tail. PLATE 4-6

7 F Distribution: F S S. Distribution of ratio of two sample variances.. Used to build confidence interval for ratio of two population variances. 3. Starts at 0 and goes to +. F(, v, v ) 4. Based number of degrees of freedom from sample. 5. Distribution is not symmetric, however: F (, v, v ) F (, v, v ) 6. Table gives critical F value which delineates area in upper tail. PLATE 4-7

8 A CONFIDENCE INTERVAL FOR A POPULATION MEAN USING A SAMPLE MEAN & STANDARD DEVIATION Lower Tail / - Confidence Region -t +t Upper Tail / µ y ± t ( /, v ) S n OR y t ( /, v) S n µ y t ( /, v) S n PLATE 4-8

9 EXAMPLE A sample of 0 circle readings has a mean of 34.5", and a standard deviation of ±.", what is the: a) 95% confidence interval for the pop. mean? b) 99% confidence interval for the pop. mean? c) would a measurement of 35.7 be acceptable for this set of data? Part a) Step : = 0.05 ( ) so / = 0.05 v = 0 - = 9 Look up critical value of t =.093 (0.05, 9) Step : µ PLATE 4-9

10 Part b: 99% CONFIDENCE INTERVAL Step : = 0.0 ( ) so / = v = 0 - = 9 Look up critical value of t (0.005, 9) =.86 Step : µ Note that the 99% confidence interval is larger than the 95%. This interval indicates that 99% of the time the population mean is between 33. and PLATE 4-0

11 TABLE D-3 Critical Values for t Distribution v PLATE 4-

12 Constructing Confidence Intervals for First Sample Set : y = 5.89 S = 8.4 Set : y = 6.4 S = 36.9 Set 3: y = 6.03 S = 9.39 Set 4: y = 3.84 S =.05 Set 5: y = 4.45 S = 0.35 Set 6: y = 6.64 S = 7.7 Set 7: y = 5.96 S = 8.65 Set 8: y = 4.0 S = 9.5 Set 9: y = 5.66 S = 8.37 Set 0: y = 6.49 S = 5.9 Construct a 90% confidence interval for µ. Does the µ of 6. lie in the interval? SET : 3.40 < µ < 8.38 SET 6: 3.6 < µ < 9.67 SET :.75 < µ < 9.73 SET 7: 4.6 < µ < 7.66 SET 3: 3.48 < µ < 8.58 SET 8:.45 < µ < 6.57 SET 4:. < µ < 6.56 SET 9: 4.74 < µ < 6.57 SET 5:.59 < µ < 6.3 SET 0: 4. < µ < 8.76 PLATE 4-

13 SELECTING A SAMPLE SIZE Specifications on a project call for 95% of all angles to be measured to within ±.." How many turnings should be measured if the standard error in a single angle is ±4"? Note: must use standard normal distribution. n t / I n PLATE 4-3

14 CAN INTERVAL BE DECREASED BEYOND REASON? Assume estimated error in angle is ±8", plot n versus size of interval. Interval Interval vs Repetitions Repetitions Note that gains in precision stabilize at 8 repetitions. (4DR) PLATE 4-4

15 A CONFIDENCE INTERVAL FOR A POPULATION VARIANCE USING A SAMPLE VARIANCE / - /,v Confidence Region /,v / v S /,v < < v S /,v NOTE: Since distribution is not symmetric, critical values must be determined for both lower and upper tails. PLATE 4-5

16 EXAMPLE A sample of 0 circle readings has a standard deviation of ±.", what is the: a) 95% confidence interval for the pop. variance? b) 99% confidence interval for the pop. variance? Part a) Step : = 0.05 ( ) so / = 0.05, v = 0 - = 9 Look up critical value of and (0.05, 9) (0.975, 9) Step :.55 9 (.) 3.85 < < 9 (.) Part b: 99% CONFIDENCE INTERVAL Step : = 0.0 ( ) so / = 0.005, v = 0 - = 9 Look up critical value of (0.005, 9) and (0.995, 9) Step :.7 9 (.) < < 9 (.) Note that the 99% confidence interval is larger than the 95%. This interval indicates that 99% of the time the population variance is between.7 and 8.9. PLATE 4-6

17 Table D- Critical Values for Distribution v PLATE 4-7

18 CONSTRUCT CONFIDENCE INTERVALS FOR POPULATION VARIANCE Recalling first data set: Set : Set : Set 3: Set 4: Set 5: Set 6: Set 7: Set 8: Set 9: Set 0: y = 5.89 y = 6.4 y = 6.03 y = 3.84 y = 4.45 y = 6.64 y = 5.96 y = 4.0 y = 5.66 y = 6.49 S = 8.4 S = 36.9 S = 9.39 S =.05 S = 0.35 S = 7.7 S = 8.65 S = 9.5 S = 8.37 S = 5.9 Construct a 90% confidence interval for. Does the 7.5 lie in the interval? of SET : 9.80 < < SET 6: 4.5 < < SET : 9.30 < < SET 7: 4.60 < < 3.38 SET 3: 0.3 < < 5.4 SET 8: 0.38 < < 5.76 SET 4:.73 < < SET 9: 4.45 < <.6 SET 5: 5.5 < < 7.97 SET 0: 8.3 < < 4.3 PLATE 4-8

19 Table D- Critical Values for Distribution v PLATE 4-9

20 TABLE D-3 Critical Values for t Distribution v PLATE 4-0

21 CONFIDENCE INTERVAL FOR RATIO OF TWO POPULATION VARIANCES - confidence region F l F u / P F l < S S < F u F ( /, v, v ) < S S < F ( /, v, v ) F ( /, v, v ) < S S < F ( /, v, v ) S S F ( /, v, v ) < < S S F ( /, v, v ) S S F ( /, v, v ) < < F ( /, v, v ) S S PLATE 4-

22 CONFIDENCE INTERVAL FOR RATIO OF TWO POPULATION VARIANCES Determine the confidence interval for the ratio of two population variances based on the two sample set variances. S S F ( /,v,v ) < < S S F ( /,v,v ) EXAMPLE: On Day, 0 EDM distance measurements result in a variance of 5 mm. On Day, additional measurements of the same distance results in a variance of 6 mm. What is the 95% confidence interval for the ratio of the population variances? Since in similar measurement conditions, the expected ratio of the variances is, i.e.,. From the constructed interval is this true? PLATE 4-

23 Solution: ) / = 0.05, ) Find critical values for F (0.05,9,0) =.84 & F (0.05,0,9) = ) Construct interval: < < ) Since is contained in interval, there is no reason to believe that at a 95% level of confidence. PLATE 4-3

24 Table D-4 (continued) Critical Values for the F Distribution, = 0.05 v v PLATE 4-4

25 Table D-4 (continued) Critical Values for the F Distribution, = 0.05 v v PLATE 4-5

26 EXAMPLE FOR CLASS PRACTICE Recalling first data set: Set : S = 8.4 Set 6: S 6 = 7.7 Set : S = 36.9 Set 7: S 7 = 8.65 Set 3: S 3 = 9.39 Set 8: S 8 = 9.5 Set 4: S 4 =.05 Set 9: S 9 = 8.37 Set 5: S 5 = 0.35 Set 0: S 0 = 5.9 Construct 95% confidence intervals for the following ratios: ) ) 3) 4) ) 6) 7) 8) ) 0) ) ) PLATE 4-6

27 Solutions to plate (0.3,.05). (0.4, 3.83) 3. (0.,3.37) 4. (0.44, 7.7) 5. (0.7,.7) 6. (0.53, 8.58) 7. (0.3, 3.80) 8. (0.55, 8.87) 9. (0.30, 4.85) 0. (0.49, 7.94).(.07, 7.5)*. (0.30, 4.98) Note that set does not contain. Thus there is reason to believe that samples and 9 are not from the same population at a 95% level of confidence. This assumptions is obviously wrong, and thus the test has resulted in an incorrect results, which can be expected 5% of the time. PLATE 4-7