- E < p. ˆ p q ˆ E = q ˆ = 1 - p ˆ = sample proportion of x failures in a sample size of n. where. x n sample proportion. population proportion

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1 1 Chapter 7 ad 8 Review for Exam Chapter 7 Estimates ad Sample Sizes 2 Defiitio Cofidece Iterval (or Iterval Estimate) a rage (or a iterval) of values used to estimate the true value of the populatio parameter Lower # < populatio parameter < Upper # As a example < p <

2 4 Cofidece Iterval for Populatio Proportio pˆ - E < p E = where z a / 2 < p ˆ + E ˆ p q ˆ Notatio for Proportios p = populatio proportio p ˆ = x sample proportio (proouced p-hat ) of x successes i a sample of size q ˆ = 1 - p ˆ = sample proportio of x failures i a sample size of 5 Roud-Off Rule for Cofidece Iterval Estimates of p Roud the cofidece iterval limits to three sigificat digits 6

3 7 Procedure for Costructig a Cofidece Iterval for p 1. Verify that the required assumptios are satisfied. (The sample is a simple radom sample, the coditios for the biomial distributio are satisfied, ad the ormal distributio ca be used to approximate the distributio of sample proportios because p 5, ad q 5 are both satisfied). 2. Refer to Table A-2 ad fid the critical value z a/2 that correspods to the desired cofidece level. 3. Evaluate the margi of error E = p q ˆˆ Procedure for Costructig a Cofidece Iterval for p 4. Usig the calculated margi of error, E ad the value of the sample proportio, p, ˆ, fid the values of p ˆ E ad p ˆ + E.. Substitute those values i the geeral format for the cofidece iterval: p ˆ E < p < p ˆ + E 5. Roud the resultig cofidece iterval limits to three sigificat digits. 8 Example: I the Chapter Problem, we oted that 829 adult Miesotas were surveyed, ad 51% of them are opposed to the use of the photo-cop for issuig traffic tickets. Use these survey results. Fid the 95% cofidece iterval estimate of the populatio proportio p. 9

4 10 Example: I the Chapter Problem, we oted that 829 adult Miesotas were surveyed, ad 51% of them are opposed to the use of the photo-cop for issuig traffic tickets. Use these survey results. First, we check for assumptios. We ote that p = , ad q = ˆ ˆ Next, we calculate the margi of error. We have foud that p = 0.51, q = = 0.49, z a/ 2 = 1.96, ad = 829. ˆ E = 1.96 E = ˆ (0.51)(0.49) 829 Example: I the Chapter Problem, we oted that 829 adult Miesotas were surveyed, ad 51% of them are opposed to the use of the photo-cop for issuig traffic tickets. Use these survey results. Fid the 95% cofidece iterval for the populatio proportio p. We substitute our values from Part a to obtai: < p < , < p < Example: I the Chapter Problem, we oted that 829 adult Miesotas were surveyed, ad 51% of them are opposed to the use of the photo-cop for issuig traffic tickets. Use these survey results. Based o the results, ca we safely coclude that the majority of adult Miesotas oppose use of the photo-cop? Based o the survey results, we are 95% cofidet that the limits of 47.6% ad 54.4% cotai the true percetage of adult Miesotas opposed to the photo-cop. The percetage of opposed adult Miesotas is likely to be ay value betwee 47.6% ad 54.4%. However, a majority requires a percetage greater tha 50%, so we caot safely coclude that the majority is opposed (because the etire cofidece iterval is ot greater tha 50%). 12

5 13 Estimatig a Populatio Mea: s Not Kow Cofidece Iterval for the Estimate of m Based o a Ukow s ad a Small Simple Radom Sample from a Normally Distributed Populatio x - E < µ < x + E where E = t s a/2 t a/2 foud i Table A-3 14 Table A-3 t Distributio Degrees of freedom.005 (oe tail).01 (two tails).01 (oe tail).02 (two tails).025 (oe tail).05 (two tails).05 (oe tail).10 (two tails).10 (oe tail).20 (two tails).25 (oe tail).50 (two tails) Large (z)

6 16 Example: A study of 12 Dodge Vipers ivolved i collisios resulted i repairs averagig $26,227 ad a stadard deviatio of $15,873. Fid the 95% iterval estimate of m, the mea repair cost for all Dodge Vipers ivolved i collisios. (The 12 cars distributio appears to be bell-shaped.) x = 26,227 s = 15,873 a = 0.05 a/2 = t a /2 = E = t a / 2 s = (2.201)(15,873) = 10, x - E < µ < x + E 26,227-10,085.3 < µ < 26, ,085.3 $16,141.7 < µ < $36,312.3 We are 95% cofidet that this iterval cotais the average cost of repairig a Dodge Viper. Ed of 7-2 ad 7-3 Determiig Sample Size Required to Estimate p ad m 17 Sample Size for Estimatig Proportio p Whe a estimate of p is kow: = ˆ z a/2 E 2 ˆ ˆ ( ) 2 p q ˆ Formula 7-2 Whe o estimate of p is kow: ( z ) Formula 7-3 = a/2 E 2 18

7 19 Example: Example:We wat to determie, with a margi of error of four percetage poits, the curret percetage of U.S. households usig . Assumig that we wat 90% cofidece i our results, how may households must we survey? A 1997 study idicates 16.9% of U.S. households used . = [z a /2 ] 2 p q E 2 ˆˆ = [1.645] 2 (0.169)(0.831) = = 238 households To be 90% cofidet that our sample percetage is withi four percetage poits of the true percetage for all households, we should radomly select ad survey 238 households. Example: Example:We wat to determie, with a margi of error of four percetage poits, the curret percetage of U.S. households usig . Assumig that we wat 90% cofidece i our results, how may households must we survey? There is o prior iformatio suggestig a possible value for the sample percetage. = [z a /2 ] 2 (0.25) E = (1.645) 2 (0.25) = = 423 households With o prior iformatio, we eed a larger sample to achieve the same results with 90% cofidece ad a error of o more tha 4%. 20 Sample Size for Estimatig Mea m E = z s a/2 (solve for by algebra) = z a/2 E s 2 Formula 7-5 z a /2 /2 = critical z score based o the desired degree of cofidece E = desired margi of error s = populatio stadard deviatio 21

8 22 Example: If we wat to estimate the mea weight of plastic discarded by households i oe week, how may households must be radomly selected to be 99% cofidet that the sample mea is withi 0.25 lb of the true populatio mea? (A previous study idicates the stadard deviatio is lb.) a = 0.01 z a/2 = E = 0.25 s = = z a/2 s = (2.575)(1.065) E 0.25 = = 121 households We would eed to radomly select 121 households ad obtai the average weight of plastic discarded i oe week. We would be 99% cofidet that this mea is withi 1/4 lb of the populatio mea. Chapter 8 Hypothesis Testig 23 Claim: Usig math symbols H 0 : Must cotai equality H 1 : Will cotai,, <, > 24

9 25 Test Statistic The test statistic is a value computed from the sample data, ad it is used i makig the decisio about the rejectio of the ull hypothesis. /\ z = p - p pq Test statistic for proportios Test Statistic The test statistic is a value computed from the sample data, ad it is used i makig the decisio about the rejectio of the ull hypothesis. t = x - µ x s Test statistic for mea 26 Test Statistic The test statistic is a value computed from the sample data, ad it is used i makig the decisio about the rejectio of the ull hypothesis. c 2 = ( 1)s2 s 2 Test statistic for stadard deviatio 27

10 28 Critical Regio Set of all values of the test statistic that would cause a rejectio of the ull hypothesis Critical Regios Critical Value Ay value that separates the critical regio (where we reject the ull hypothesis) from the values of the test statistic that do ot lead to a rejectio of the ull hypothesis Reject H 0 Fail to reject H 0 Critical Value ( z score ) 29 Two-tailed, Right-tailed, tailed, Left-tailed tailed Tests The tails i a distributio are the extreme regios bouded by critical values. 30

11 31 Decisio Criterio Traditioal method: Reject H 0 if the test statistic falls withi the critical regio. Fail to reject H 0 if the test statistic does ot fall withi the critical regio. Wordig of Fial Coclusio Figure Comprehesive Hypothesis Test 33

12 34 Example: It was foud that 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted i hospitalizatio of the drivers. Usig the 0.01 sigificace level, test the claim that the air bag hospitalizatio is lower tha the 7.8% rate for cars with automatic tic safety belts. Claim: p < p = 46 / 821 = reject H 0 H 0 : p = H 1 : p < a = 0.01 p = z = p - p pq (0.078 )(0.922) 821 z = =» z = p = There is sufficiet evidece to support claim that the air bag hospitalizatio rate is lower tha the 7.8% rate for automatic safety belts. 8-5 Testig a Claim about a Mea: s Not Kow 35 Example: Seve axial load scores are listed below. At the 0.01 level of sigificace, test the claim that this sample comes from a populatio with a mea that is greater tha 165 lbs = 7 df = 6 x = lb s = 27.6 lb Claim: Claim: µ > 165 lb H 0 : µ = 165 lb H 1 : µ > 165 lb (right tailed test) 36

13 37 a = t = t = x - µ t = x s = = Reject H o Example: Seve axial load scores are listed below. At the 0.01 level of sigificace, test the claim that this sample comes from a populatio with a mea that is greater tha 165 lbs Fial coclusio: There is sufficiet evidece to support the claim that the sample comes from a populatio with a mea greater tha 165 lbs. Reject Claim: µ > 165 lb H 0 : µ = 165 lb H 1 : µ > 165 lb (right tailed test) Testig a Claim about a Stadard Deviatio or Variace 39

14 40 Chi-Square Distributio Test Statistic X 2 = ( - 1) s 2 s 2 s 2 s 2 = sample size = sample variace = populatio variace (give i ull hypothesis) Critical Values ad P-values for Chi-Square Distributio Foud i Table A-4 Degrees of freedom = -1 Based o cumulative areas from the RIGHT 41 Table A-4: Critical values are foud by determiig the area to the RIGHT of the critical value df = 80 a = 0.05 a/2 =

15 43 Example: Aircraft altimeters have measurig errors with a stadard deviatio of 43.7 ft. With ew productio equipmet, 81 altimeters ers measure errors with a stadard deviatio of 52.3 ft. Use the sigificace level to test the claim that the ew altimeters have a stadard deviatio differet from the old value of 43.7 ft. Claim: s 43.7 H 0 : s = 43.7 H 1 : s a = a/2 = = 81 df = 80 Table A x 2 = ( -1)s 2 (81-1) (52.3) = 2» s Reject H x 2 = Example: Aircraft altimeters have measurig errors with a stadard deviatio of 43.7 ft. With ew productio equipmet, 81 altimeters ers measure errors with a stadard deviatio of 52.3 ft. Use the sigificace level to test the claim that the ew altimeters have a stadard deviatio differet from the old value of 43.7 ft. SUPPORT REJECT Claim: s 43.7 H 0 : s = 43.7 H 1 : s 43.7 The ew productio method appears to be worse tha the old method. The data supports that there is more variatio i the error readigs tha before. 45

16 46 Table 8-3 Hypothesis Tests Parameter Coditios Distributio ad Test Statistic Critical ad P-values Proportio p =5 ad q =5 Normal: z p = ˆ p p q Table A-2 Mea σ ot kow ad ormally distributed or =30 Studet t: t = X s µ X Table A-3 Stadard Deviatio or Variace Populatio ormally distributed Chi-Square: x 2 ( 1) = 2 σ 2 s Table A-4 47