Statistical Inference About Means and Proportions With Two Populations

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Departmet of Quatitative Methods & Iformatio Systems Itroductio to Busiess Statistics QM 220 Chapter 10 Statistical Iferece About Meas ad Proportios With Two Populatios Fall 2010 Dr. Mohammad Zaial 1

Chapter 10 Statistical Iferece About Meas ad Proportios With Two Populatios Ifereces About the Differece Betwee Two Populatio Meas: s 1 ad s 2 Kow Ifereces About the Differece Betwee Two Populatio Meas: s 1 ad s 2 Ukow Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Ifereces About the Differece Betwee Two Populatio Proportios 2

Why do we eed to compare two populatios? You have just received a job offer i two differet coutries with the same salary. You wat to ivest your moey ito two differet stock markets. A pharmaceutical compay just aouced a ew pai killer that is better tha a old oe. A bak maager itroduced a ew servig policy that reduces the waitig time. You wat to decide betwee two cars 3

Ifereces About the Differece Betwee Two Populatio Meas: s 1 ad s 2 Kow Iterval Estimatio of m 1 m 2 Hypothesis Tests About m 1 m 2 4

Estimatig the Differece Betwee Two Populatio Meas Let m 1 equal the mea of populatio 1 ad m 2 equal the mea of populatio 2. The differece betwee the two populatio meas is m 1 - m 2. To estimate m 1 - m 2, we will select a simple radom sample of size 1 from populatio 1 ad a simple radom sample of size 2 from populatio 2. x 1 2 Let equal the mea of sample 1 ad x 2 equal the mea of sample 2. The poit estimator of the differece betwee the meas of the populatios 1 ad 2 is. x x 5

Samplig Distributio of x Expected Value x x 1 x 2 Stadard Deviatio (Stadard Error) s s s x 1 x 2 2 2 where: s 1 = stadard deviatio of populatio 1 s 2 = stadard deviatio of populatio 2 1 = sample size from populatio 1 2 = sample size from populatio 2 6

Iterval Estimate Iterval Estimatio of m 1 - m 2 : s 1 ad s 2 Kow s s x 1 x 2 z / 2 2 2 where: 1 - is the cofidece coefficiet 7

Iterval Estimatio of m 1 - m 2 : s 1 ad s 2 Kow Example: Par Products Par Products is a maufacturer of golf equipmet ad has developed a ew ball that has bee desiged to provide extra distace. I a test of drivig distace usig a mechaical drivig device, a sample of Par golf balls was compared with a sample of golf balls made by Rap Ltd, a competitor. The sample statistics appear o the ext slide. 8

Iterval Estimatio of m 1 - m 2 : s 1 ad s 2 Kow Example: Par Products Sample Size Sample Mea Sample #1 Par Sample #2 Rap 120 balls 80 balls 275 metres 258 metres Based o data from previous drivig distace tests, the two populatio stadard deviatios are kow with s 1 = 15 metres ad s 2 = 20 metres. Let us costruct a 95% cofidece iterval estimate of the differece betwee the mea drivig distaces of the two brads of golf ball. 9

Estimatig the Differece Betwee Two Populatio Meas Populatio 1 Par Products Golf Balls m 1 = mea drivig distace of Par golf balls m 1 m 2 = differece betwee the mea distaces Populatio 2 Rap Ltd. Golf Balls m 2 = mea drivig distace of Rap golf balls Simple radom sample of 1 Par golf balls x 1 = sample mea distace for the Par golf balls Simple radom sample of 2 Rap golf balls x 2 = sample mea distace for the Rap golf balls x 1 - x 2 = Poit Estimate of m 1 m 2 10

Poit Estimate of m 1 - m 2 Poit estimate of m 1 m 2 = x 1 - x 2 where: = 275 258 = 17 metres m 1 = mea distace for the populatio of Par Products golf balls m 2 = mea distace for the populatio of Rap Ltd golf balls 11

Iterval Estimatio of m 1 - m 2 : s 1 ad s 2 Kow s s (15) (20) x 1 - x 2 ± z a / 2 + = 17 ± 1.96 + 120 80 2 2 2 2 17 + 5.14 or 11.86 metres to 22.14 metres We are 95% cofidet that the differece betwee the mea drivig distaces of Par Products balls ad Rap Ltd balls is 11.86 to 22.14 metres. 12

Hypotheses Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow H H : : m m D m m D 0 0 1 0 H H : : m m D m m D 0 0 1 0 H H : : m m D m m D 0 0 1 0 Lower tail Upper tail Two-tailed Test Statistic z ( x x ) D 0 s s 2 2 13

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow Example: Par Products Ca we coclude, usig = 0.01, that the mea drivig distace of Par golf balls is greater tha the mea drivig distace of Rap golf balls? 14

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow p Value ad Critical Value Approaches 1. Develop the hypotheses. H 0 : m 1 - m 2 < 0 H 1 : m 1 - m 2 > 0 where: m 1 = mea distace for the populatio of Par golf balls m 2 = mea distace for the populatio of Rap golf balls 2. Specify the level of sigificace. = 0.01 15

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow p Value ad Critical Value Approaches 3. Compute the value of the test statistic. z ( x x ) D 0 s s 2 2 (235 218) 0 17 z 2 2 (15) (20) 2.62 120 80 6.49 16

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow p Value Approach 4. Compute the p value. For z = 6.49, the p value < 0.0001. 5. Determie whether to reject H 0. Because p value < = 0.01, we reject H 0. At the 0.01 level of sigificace, the sample evidece idicates the mea drivig distace of Par golf balls is greater tha the mea drivig distace of Rap golf balls. 17

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Kow Critical Value Approach 4. Determie the critical value ad rejectio rule. For = 0.01, z 0.01 = 2.33 Reject H 0 if z > 2.33 5. Determie whether to reject H 0. Because z = 6.49 > 2.33, we reject H 0. The sample evidece idicates the mea drivig distace of Par golf balls is greater tha the mea drivig distace of Rap golf balls. 18

Ifereces About the Differece Betwee Two Populatio Meas: s 1 ad s 2 Ukow & Uequal Iterval Estimatio of m 1 m 2 Hypothesis Tests About m 1 m 2 Whe s 1 ad s 2 are ukow, we shall: use the sample stadard deviatios s 1 ad s 2 as estimates of s 1 ad s 2, ad replace z /2 with t /2. 19

Iterval Estimatio of m 1 - m 2 : s 1 ad s 2 Ukow & Uequal Iterval Estimate s x 1 x 2 t / 2 s 2 2 where the degrees of freedom for t /2 are: df s s 2 2 2 2 2 2 2 1 s 2 1 s 1 1 1 1 2 1 2 20

Differece Betwee Two Populatio Meas: s 1 ad s 2 Ukow & Uequal Example: Specific Motors Specific Motors has developed a ew car kow as the E car. 24 E cars ad 28 J cars (from Japa) were road tested to compare kilometres-per-litre (kpl) performace. The sample statistics are show o the ext slide. 21

Differece Betwee Two Populatio Meas: s 1 ad s 2 Ukow & Uequal Example: Specific Motors Sample #1 E Cars Sample #2 J Cars 24 cars 28 cars 19.8 kpl 17.3 kpl 2.56 kpl 1.81 kpl Sample Size Sample Mea Sample Std. Dev. 22

Differece Betwee Two Populatio Meas: s 1 ad s 2 Ukow & Uequal Example: Specific Motors Let us costruct a 90% cofidece iterval estimate of the differece betwee the kpl performaces of the two models of car. 23

Poit Estimate of m 1 m 2 Poit estimate of m 1 m 2 = where: = 19.8-17.3 = 2.5 kpl m 1 = mea kilometres per litre for the populatio of E cars m 2 = mea kilometres per litre for the populatio of J cars x - x 24

Iterval Estimatio of m 1 m 2: s 1 ad s 2 Ukow & Uequal The degrees of freedom for t /2 are: df 2 2 (2.56) (1.81) 24 28 24.07 24 2 2 2 2 1 (2.56) 1 (1.81) 24 1 24 28 1 28 2 With /2 = 0.05 ad df = 24, t /2 = 1.711 25

Iterval Estimatio of m 1 m 2: s 1 ad s 2 Ukow & Uequal s s (2.56) (1.81) x 1 x 2 t / 2 19.8 17.3 1.711 24 28 2 2 2 2 1.5 + 1.069 or 1.431 to 3.569 kpl We are 90% cofidet that the differece betwee the kilometres-per-litre performaces of E cars ad J cars is 1.431 to 3.569 kpl. 26

Hypotheses Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal H H : : m m D m m D 0 0 1 0 H H : m m D : m m D 0 0 1 0 H H : : m m D m m D 0 0 1 0 Lower tail Upper tail Two-tailed Test Statistic t ( x x ) D 0 s s 2 2 27

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal Example: Specific Motors Ca we coclude, usig a 0.05 level of sigificace, that the kilometres-per-litre (kpl) performace of E cars is greater tha the kilometres- per-litre performace of J cars? 28

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal Critical Value Approaches 1. Develop the hypotheses. H 0 : m 1 - m 2 < 0 H 1 : m 1 - m 2 > 0 where: m 1 = mea kpl for the populatio of E cars m 2 = mea kpl for the populatio of J cars 29

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal Critical Value Approaches 2. Specify the level of sigificace. = 0.05 3. Compute the value of the test statistic. t ( x x ) D (19.8 17.3) 0 0 2 2 2 2 s 1 s 2 (2.56) (1.81) 24 28 4.003 30

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal Critical Value Approach 4. Determie the critical value ad rejectio rule. The degrees of freedom for t are: df 2 2 (2.56) (1.81) 24 28 24.07 24 2 2 2 2 1 (2.56) 1 (1.81) 24 1 24 28 1 28 2 For = 0.05 ad df = 24, t 0.05 = 1.711 Reject H 0 if t > 1.711 31

Hypothesis Tests About m 1 m 2: s 1 ad s 2 Ukow & Uequal Critical Value Approach 5. Determie whether to reject H 0. Because 4.003 > 1.711, we reject H 0. At the 5% sigificace level, we coclude that the kilometres-per-litre (kpl) performace of E cars is greater tha the kilometres-per-litre performace of J cars. 32

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples With a matched-samples desig each sampled item provides a pair of data values. This desig ofte leads to a smaller samplig error tha the idepedet-samples desig because variatio betwee sampled items is elimiated as a source of samplig error. 33

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Example: Express Deliveries A Lodo-based firm has documets that must be quickly distributed to brach offices throughout Europe. The firm must decide betwee two delivery services, UPX (Uited Parcel Express) ad INTEX (Iteratioal Express), to trasport its documets. 34

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Example: Express Deliveries I testig the delivery times of the two services, the firm set two reports to a radom sample of its Europea offices with oe report carried by UPX ad the other report carried by INTEX. Do the data o the ext slide idicate a differece i mea delivery times for the two services? Use a 0.05 level of sigificace. 35

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Europea Office Stockholm Copehage Hamburg Brussels Amsterdam Berli Lisbo Madrid Paris Frakfurt Delivery Time (Hours) UPX INTEX Differece 32 30 19 16 15 18 14 10 7 16 25 24 15 15 13 15 15 8 9 11 7 6 4 1 2 3-1 2-2 5 36

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Critical Value Approaches 1. Develop the hypotheses. H 0 : m d = 0 H 1 : m d Let m d = the mea of the differece values for the two delivery services for the populatio of district offices 37

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Critical Value Approaches 2. Specify the level of sigificace. = 0.05 3. Compute the value of the test statistic. d d i (7 6... 5) 2.7 10 s d t 2 ( d i d ) 76. 1 1 9 d m d 2.7 0 s 2.9 10 d 2. 9 2.94 38

Ifereces About the Differece Betwee Two Populatio Meas: Matched Samples Critical Value Approach 4. Determie the critical value ad rejectio rule. For = 0.05 ad df = 9, t 0.025 = 2.262. Reject H 0 if t > 2.262 5. Determie whether to reject H 0. Because t = 2.94 > 2.262, we reject H 0. At the 5% sigificace level, we coclude that there is a differece i mea delivery times for the two services. 39

Ifereces About the Differece Betwee Two Populatio Proportios Iterval Estimatio of p 1 - p 2 Hypothesis Tests About p 1 - p 2 Expected Value E ( p ˆ p ˆ ) p p Stadard Deviatio (Stadard Error) s p 1(1 p1) p2(1 p2 ) p 1 p 2 where: 1 = size of sample take from populatio 1 2 = size of sample take from populatio 2 40

Samplig Distributio of p ˆ 1 p ˆ 2 If the sample sizes are large, the samplig distributio of p 1 p 2 ca be approximated by a ormal probability distributio. The sample sizes are sufficietly large if all of these coditios are met: 1 p 1 > 5 1 (1 - p 1 ) > 5 2 p 2 > 5 2 (1 - p 2 ) > 5 41

Samplig Distributio of p ˆ 1 pˆ 2 s p 1(1 p1) p2(1 p2 ) p 1 p 2 p 1 p 2 p ˆ p ˆ 42

Iterval Estimatio of p 1 - p 2 Iterval Estimate p ˆ 1 (1 pˆ 1 ) pˆ 2 (1 pˆ 2 ) p ˆ 1 pˆ 2 z /2 43

Iterval Estimatio of p 1 - p 2 Example: Market Research Associates Market Research Associates is coductig research to evaluate the effectiveess of a cliet s ew advertisig campaig. Before the ew campaig bega, a telephoe survey of 150 households i the test market area showed 60 households aware of the cliet s product. The ew campaig has bee iitiated with TV ad ewspaper advertisemets ruig for three weeks. 44

Iterval Estimatio of p 1 - p 2 Example: Market Research Associates A survey coducted immediately after the ew campaig showed 120 of 250 households aware of the cliet s product. Do the data support the positio that the advertisig campaig has provided a icreased awareess of the cliet s product? 45

Poit Estimator of the Differece Betwee Two Populatio Proportios p 1 = proportio of the populatio of households aware of the product after the ew campaig p 2 = proportio of the populatio of households aware of the product before the ew campaig ˆp 1 ˆp 2 = sample proportio of households aware of the product after the ew campaig = sample proportio of households aware of the product before the ew campaig 120 60 p ˆ 1 pˆ 2 0.48 0.40 0.08 250 150 46

Iterval Estimatio of p 1 - p 2 For = 0.05, z 0.025 = 1.96 0.48 0.40 1.96 0.48(0.52) 0.40(0.60) 250 150 0.08 + 1.96(0.0510) 0.08 + 0.10 Hece, the 95% cofidece iterval for the differece i before ad after awareess of the product is -0.02 to +0.18. 47

Hypothesis Tests about p 1 - p 2 Hypotheses We focus o tests ivolvig o differece betwee the two populatio proportios (i.e. p 1 = p 2 ) H 0 : p 1 p 2 0 H : p p 0 1 H 0 : p 1 - p 2 < 0 H 0 : p 1 p 2 0 H a : p 1 - p 2 > 0 1 H 0 : p 1 p 2 0 H : p p 0 1 Lower tail Upper tail Two-tailed 48

Hypothesis Tests about p 1 - p 2 Pooled Estimate of Stadard Error of p 1 p 2 s 1 1 ˆ 1 ˆ ˆ ˆ p p p(1 p) 2 1 2 where: p ˆ p ˆ p ˆ 1 2 49

Hypothesis Tests about p 1 - p 2 Test Statistic z ( p ˆ p ˆ ) 1 1 p ˆ (1 p ˆ ) 50

Hypothesis Tests about p 1 - p 2 Example: Market Research Associates Ca we coclude, usig a 0.05 level of sigificace, that the proportio of households aware of the cliet s product icreased after the ew advertisig campaig? 51

Hypothesis Tests about p 1 - p 2 p-value ad Critical Value Approaches 1. Develop the hypotheses. H 0 : p 1 - p 2 < 0 H 1 : p 1 - p 2 > 0 p 1 = proportio of the populatio of households aware of the product after the ew campaig p 2 = proportio of the populatio of households aware of the product before the ew campaig 52

Hypothesis Tests about p 1 - p 2 p -Value ad Critical Value Approaches 2. Specify the level of sigificace. = 0.05 3. Compute the value of the test statistic. 250(0.48) 150(0.40) 180 p ˆ 0.45 250 150 400 s 1 1 p ˆ 1 p ˆ 0.45(0.55)( ) 0.0514 2 250 150 (0.48 0.40) 0 0.08 z 1.56 0.0514 0.0514 53

Hypothesis Tests about p 1 - p 2 p Value Approach 4. Compute the p value. For z = 1.56, the p value = 0.0594 5. Determie whether to reject H 0. Because p value > = 0.05, we caot reject H 0. We caot coclude that the proportio of households aware of the cliet s product icreased after the ew campaig. 54

Hypothesis Tests about p 1 - p 2 Critical Value Approach 4. Determie the critical value ad rejectio rule. For = 0.05, z 0.05 = 1.645 Reject H 0 if z > 1.645 5. Determie whether to reject H 0. Because 1.56 < 1.645, we caot reject H 0. We caot coclude that the proportio of households aware of the cliet s product icreased after the ew campaig. 55

Ed of Chapter 10 56