# Lecture 5 -

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Unformatted text preview: MGCR 271! !"#\$%&##'()*+#+,#' !"#\$%&'(\$)\$'*\$% -*.%*)/'0*\$12*%*)/*%' Inference on Proportions! Chapter 8! 3&)' "#' #"445#&' 25"' )5##' *' ,5\$%' 67' +.&#' *%1' \$)' )"8%#' )*\$9#' :' +.&#;' <*%' 25"' ,5%,9"1&')/*)'\$)'\$#'%5)'*'=*\$8',5\$%;'>/2'58' ?/2'%5);' >/*)' \$#' )/&' [email protected]*@\$9\$)2' )/*)' )/&8&' ?5"91' @&'AB:')*\$9#'\$%'%B67')5##&#'5='*',5\$%;' Sampling Distribution for Sample Proportion is Approximately Normal! C/&'#*.49\$%D'1\$#)[email protected]"+5%'5='*'#*.49&'485458+5%'''''''\$#'*4485A\$.*)&92' %58.*9'E%58.*9'*4485A\$.*+5%'5='*'@\$%5.\$*9'1\$#)[email protected]"+5%F'?/&%')/&' #*.49&'#\$G&'\$#'9*8D&'&%5"D/H' Z= Z ∼ N (0, 1) z= hkjhjk ˆ q p−p0 ˆ qP −p p(1−p) n MGCR 271: Business Statistics Sample Questions morning. Pete, an engineering student at Purdue, thought that since engineering students are 1. A local news station reported that 72% of all people push the snooze button at least once before waking pup −p0 ) 0 (1 in the n usually up late studying, a higher percentage of engineers would push the snooze button. He decides to take a sample of 50 engineering students and found that 39 said they push the snooze button each morning. Is the true percentage of people who push the snooze button at least once in the morning (use α = 0.10 ) signiﬁcantly higher than the news station’s report? 1. gfhgfgh (a) Identify the population(s), sample(s). (a) State the null and alternative hypotheses. Use only letters and math(b) ematical null and alternate hypotheses for this test. Do not use words. State the symbols to write the hypotheses. (c) Identify the entity, property, and the standard error parameter and (b) Identifythe parameter, its statisticpopulation, sample, of the statistic. statis(d) tic for the test statistic, its distribution and (a). value(s). Identify the hypothesis being tested in critical (c) Identify the values of the standard error Distribution (e) Calculate the Test Statistic and its and test statistic. (d) Computethe p-Value. (f) Calculate the Critical Test Statistic(s) (g) State your the Test Score of the story. (e) Compute conclusion in termsand the p-Value (h) On the curve below diagrams of label the p0 , the p, of the Value for the hypothesis test ˆ (f) Draw and label insert and clearlythe distributionand the p−test statistic. in parts indicate (d) critical region(s), the critical test statistic(s), Clearly (a) through theabove. (i) the test score. Shade the area corresponding to the p-value. Construct a 90% conﬁdence interval for the proportion of engineering students who push the snooze button. (g) State your conclusion (whether or not you reject the null hypothesis) and state why. 2. An environmental health professor conducted a study to see whether fast-food workers wearing gloves actually lowers the chance that customers will come down with food poisoning. The scientists purchased 371 tortillas from several local fast-food restaurants, noting whether the workers were wearing gloves or not. 190 of the tortillas came from bare-hands restaurants; 181 of the tortillas came from glove-wearing I%&'(*.49&'J85458+5%' K Formulate Hypothesis! Parameter! Null Value of Parameter! H 0 : p = p0 H a : p = p0 6 Identify Test Statistic…! ˆ P p Test Statistic! ˆ P p ˆ P p(1−p) Statistic ! p n Z= ˆp(1−p) P −pn q p(1−p) n q p(1−p) ˆn P −p Parameter! p(1−p) n ˆ P p Std. Error! p(1−p) n ˆ qP −p p(1−p) ˆ P −p q p(1−p) n p(1−p) n 6 …and its Distribution! p(1−p) n !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Z= Z ∼ N (0, 1) ˆ qP −p 0.3 ˆ P p population group ! 0.2 a b c ! ! 0.1 p(1−p) n 0 !3 !2 Z= !1 x ˆ P −p q 1 2 3 p(1−p) n L p(1−p) n Compute Test Score! Z= Test Score! Z ∼ N (0, 1) z= ˆ q p−p0 p0 (1−p0 ) n Statistic (Observed Value) ! ˆ P −p q p(1−p) n Parameter (Null Value)! Std. Error! One-Proportion z-Test! p −p ˆ ˆ 1 2 z= q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − µ2 = d0 0.2 Ha : µ1 − µ2 = d0 p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! N (0, 1) Fail to RejectFail to Reject H0 H0 H0 : µ1 − µ2 = d0 Ha : µ1 − µ2 = d0 ! ! group group a b b c c ! H 0 : p = p0 H a : p = p0 α 2 x1 − x2 ¯ ¯ p 2 z∗ −z ∗ !2 !1 0 1 x p0 z∗ z = −z ∗ 2 3 p ˆ ˆ q p−p0 p0 (1−p0 ) n p p(1−p) n (j) If you believed that the standard deviations for the two populations were identical, how would you modify your test. Speciﬁcally, indicate the test statistic you would use and compute its value. Z= 3. An environmental health professor conducted a study to see whether fast-food workers wearing gloves ˆ qP −p actually lowers p(1−p) n the chance that customers will come down with food poisoning. The scientists purchased Z ∼ N (0, 1)190 of the tortillas came from bare-hands restaurants; 181 of the tortillas came from glove-wearing not. z= ˆ restaurants. q p−p0 p0 (1−p0 ) n 371 tortillas from several local fast-food restaurants, noting whether the workers were wearing gloves or The scientists then tested the tortillas purchased for microbe growth. They found that the bare-hands restaurants’ tortillas gave rise to microbe growth on 18 tortillas, and the glove-wearing tortillas’ microbe growth signiﬁcantly lower than the bare-hands restaurants’ microbe growth at the hkjhjk restaurants’ tortillas gave rise to microbe growth only on 8 tortillas. Is the glove-wearing restaurants’ 1. gfhgfgh 5% signiﬁcance leve.? (a) State thethe population(s), sample(s) (a) Identify null and alternative hypotheses. Use only letters and mathematical symbols to write the hypotheses. Do not use words. (b) Identify the entity, property, population, sample, parameter and statis(c) Identify the test statistic, its distribution and critical value(s). tic for the hypothesis being tested in (a). (d) Calculate the values of the (c) Identify the Test Statisticstandard error and the test statistic and its Distribution (e) Calculate the Critical (d) Compute the p−Value. Test Statistic(s) (f) State your conclusion in terms the p-Value (e) Compute the Test Score and of the story. (g) On the curve below insert and the label the values the test statistic. (f) Draw and label diagrams ofclearlydistribution of of the parameter, statistic, test statistic, critical indicate the critical region(s), the region(s) and the p−Value. Clearly value(s) of the test statistic, the rejectioncritical test statistic(s), (h)the test score. Shade the area correspondingerence in the number of tortillas with microbe Construct a 95% conﬁdence interval for the diﬀ to the p-value. (b) Identify the parameter, its statistic and the standard error of the statistic. (g) State your conclusiontypes of restaurants. reject the null hypothesis) growth across the two (whether or not you and state why. C?5'(*.49&'J85458+5%#' H0 : µd = 0 K Ha : µd > 0 Formulate Hypothesis! T= D −µd √ sd / n Parameter! Null Value of Parameter! ¯ ¯ D = X1 − X2 H 0 : p 1 − p2 = 0 H a : p 1 − p2 > 0 2 = d0 µ1 − µ2 d0 p0 X 1 − X 2 ¯ ¯ p 1 − p2 d0 1 2 p ˆ Sp 1 n1 + 1 n2 x −x ¯ ¯ s +, Identify Test Statistic…! proportion z-test H :p=p p 1 n1 1 n2 − p26 = d0 ˆ ˆ P1 − P2 p1 − p 2 ˆ ˆ p0 (1−p0 ) t(df ) n ˆ ˆ ˆ ˆ P1 (1−P1 ) P2 (1−P2 ) + n1 n2 0 0 prop. z-test, p Unpooled nequal Variances) 2 − p2 = d 0 = d0 p 1 − p2 ˆ P p ˆ p0 H 0 : p1 − p2 = d Statistic ! p1 − p ˆ ˆ ˆ ˆ P1 − P2 p1 (1−p1 ) ˆ ˆ 2 (n1 −1)s2 +(n2 −1)sn1 n1 +n2 −2 p(1−p) n 0 1 2 sp = p p0 + 1 n1 p df = n1 + n2 − 2 p2 (1−p2 ) ˆ ˆ n2 ˆ P ˆ ˆ P (1 − P ) p(1 − p) ˆ ˆ p= ˆ 1 n2 p ˆ 1 se when d0 = 0 Test Statistic! P − P p1 − p2 d0 ˆ1 ˆ2 2 ˆ ˆ P1 − P2 p1 − p 2 ˆ ˆ p0 (1−p0 ) n ˆ ˆ ˆ ˆ P1 (1−P1 ) + P2 (1−P2 ) n2 2 n1 Parameter! n1 p1 +n2 p2 n1 +n2 p1 − + 2 1 nN (0, 1) ˆ 2 P − d0 1 n1 + 1 N (0, 1) n2 prop. z-test,dPooled 0 2 H p1 − p 2 0 ˆ ˆ ˆ1 p ˆ ˆ ˆ : p1 −pp(1−=)d0 p2 (1−p2 ) p1 − p2 21 1 + n n2 = d0 p 1 − p2 qual Variances) d0 2 ˆ ˆ P (1 − P ) p(1 − p) ˆ ˆ p= ˆ ˆ P1 − N (0, 1) p1 − ˆ se when d0 = 0 Std. Error! 1 n1 1 n1 + + 1 n2 d0 2 p1 − ˆ n1 p1 +n2 p2 n1 +n2 p(1−p) n 6 …and its Distribution! p(1−p) n !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! Z= Z ∼ N (0, 1) ˆ qP −p 0.3 population group ! 0.2 a b c ! ! 0.1 !3 !2 !1 0 1 2 3 x L Compute Test Score! Test Score! Statistic (Observed Value) ! Parameter (Null Value)! z= p= ˆ n1 p1 +n2 p2 ˆ ˆ n1 +n2 Std. Error! q p(1−p) ˆ ˆ p(1−p) ˆ ˆ +n n1 2 (p1 −p2 )−d0 ˆˆ z= p= ˆ n1 p1 +n2 p2 ˆ ˆ n1 +n2 q p(1−p) ˆ ˆ p(1−p) ˆ ˆ +n n1 2 (p1 −p2 )−d0 ˆˆ Two-Proportion pz-Test, Pooled! p p −ˆ ˆ 1 2 ˆ z= P q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − 0.2 Ha : µ1 − µ2 = d0 hkjhjk p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! p(1−p) n N (0, 1) ˆ qP −p p(1−p) Fail to RejectFail to Reject H0 H0 n Z= Z ∼ N (0, 1) µ2 = q p−p0 = d0 ˆ z p0 (1−p0 ) n ( ˆ1 − H0 µ1(1− ˆˆ2 = −0 µ p d dˆ z = :q pp− pp2 )−(10 p) ˆ ) ˆ H 0 : p 1 − p2 = 0 1. gfhgfgh H a : p 1 − p2 = 0 α 2 Ha :nµp1−nµp2= d0 1 1 ˆ + 22 p = n1 +n2 ˆ ˆ x1 − x2 ¯ ¯ ! ! group group a b b c c ! n1 + n2 z ∗ p 2 −z !2 ∗ !1 H0 : 2 (b) Identify the entity, property, population, samp −z ∗ p1 − pn1 ˆ1 +n2 tested in (a). ˆ ˆ2 p1 − p2 = d0 tic for the hypothesispbeing p2 ˆ (a) State the null and alternative )hypotheses. Use ( p1 −p ˆ 0 (p1 − ˆ2 ) −d ˆ 0 1 2∗ zz=3 qq write pˆ2 −2d(10 p2 ) = to1 (1(1−1p)thepp(1− ˆ ) ˆ x ematicalz symbols p p −p ˆ)+ ˆhypotheses. Do ˆˆ ˆ −p ˆ n1 n 1 + n2 n H :p −p =d 1 2 (c) Identify the Test Statistic and its Distribution p= ˆ n +n Inference on Proportions! Summary! Response! Variable! Explanatory! Variable! K sample! None! 2 samples! Independent! >2 samples! Pooled! 2-prop. z-test, Pooled! Two Way Tables! O Unpooled! 2-prop. z-test, Unpooled! N K-prop. z-test! M Categ.! Categ.! : KK' Quant.! Logistic Regression! M One-Proportion z-Test! p −p ˆ ˆ 1 2 z= q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − µ2 = d0 0.2 Ha : µ1 − µ2 = d0 p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! N (0, 1) Fail to RejectFail to Reject H0 H0 H0 : µ1 − µ2 = d0 Ha : µ1 − µ2 = d0 ! ! group group a b b c c ! H 0 : p = p0 H a : p = p0 α 2 x1 − x2 ¯ ¯ p 2 z∗ −z ∗ !2 !1 0 1 x p0 z∗ z = −z ∗ 2 3 p ˆ ˆ q p−p0 p0 (1−p0 ) n Response! Variable! Explanatory! Variable! K sample! None! 2 samples! Independent! >2 samples! Pooled! 2-prop. z-test, Pooled! Two Way Tables! O Unpooled! 2-prop. z-test, Unpooled! N K-prop. z-test! M Categ.! Categ.! : KK' Quant.! Logistic Regression! N Two-Proportion − p z-Test, Unpooled! p ˆ ˆ 1 2 z= q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − µ2 = d0 0.2 Ha : µ1 − µ2 = d0 p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! N (0, 1) Fail to RejectFail to Reject H0 H0 H0 : µ1 − µ2 = d0 Ha : µ1 − µ2 = d0 ! ! group group a b b c c ! H 0 : p1 − p 2 = d 0 Ha : p1 − p2 = d 0 α 2 x1 − x2 ¯ ¯ p 2 p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ p1 (1−p1+ 2 2 (1−2 2 ) ˆ ˆ) p ˆ p ˆ n1 + nn 22 n1 z∗ −z !2 ∗ !1 0 1 x H 0 : p1 − p 2 = d 0 H :p −p =d z −z ∗ 2 (ˆ ˆˆ ∗z z = q (p1 −−2 )−d0 0 =3 q ˆ p1 p p2 )−d p1 − p 2 ˆ ˆ Response! Variable! Explanatory! Variable! K sample! None! 2 samples! Independent! >2 samples! Pooled! 2-prop. z-test, Pooled! Two Way Tables! O Unpooled! 2-prop. z-test, Unpooled! N K-prop. z-test! M Categ.! Categ.! : KK' Quant.! Logistic Regression! O Two-Proportion pz-Test, Pooled! p p −ˆ ˆ 1 2 ˆ z= P q p1 (1−p1 ) ˆ ˆ p (1−p ) ˆ ˆ + 2n 2 n1 2 (p1 −p2 )−d0 ˆˆ Fail to Reject H0 0.3 population H0 : µ1 − 0.2 Ha : µ1 − µ2 = d0 hkjhjk p 2 0.1 x1 − x2 ¯ ¯ α 2 !3 0.0 !! !! ! !!! !! !!! !! !! !!! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! !! ! !! ! !! !! ! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! !! ! !! ! !! !! !! !! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!!! !!!! !!!! !!! !!! !!! !!! !!!!! !! !!! !!!!!!! !!! !!!!!!! !!! !!!!!!! !!!!!! !! !!!!!!! ! !!!!!!! ! !!!!!!!! !!!!!!!! p(1−p) n N (0, 1) ˆ qP −p p(1−p) Fail to RejectFail to Reject H0 H0 n Z= Z ∼ N (0, 1) µ2 = q p−p0 = d0 ˆ z p0 (1−p0 ) n ( ˆ1 − H0 µ1(1− ˆˆ2 = −0 µ p d dˆ z = :q pp− pp2 )−(10 p) ˆ ) ˆ H 0 : p 1 − p2 = 0 1. gfhgfgh H a : p 1 − p2 = 0 α 2 Ha :nµp1−nµp2= d0 1 1 ˆ + 22 p = n1 +n2 ˆ ˆ x1 − x2 ¯ ¯ ! ! group group a b b c c ! n1 + n2 z ∗ p 2 −z !2 ∗ !1 H0 : 2 (b) Identify the entity, property, population, samp −z ∗ p1 − pn1 ˆ1 +n2 tested in (a). ˆ ˆ2 p1 − p2 = d0 tic for the hypothesispbeing p2 ˆ (a) State the null and alternative )hypotheses. Use ( p1 −p ˆ 0 (p1 − ˆ2 ) −d ˆ 0 1 2∗ zz=3 qq write pˆ2 −2d(10 p2 ) = to1 (1(1−1p)thepp(1− ˆ ) ˆ x ematicalz symbols p p −p ˆ)+ ˆhypotheses. Do ˆˆ ˆ −p ˆ n1 n 1 + n2 n H :p −p =d 1 2 (c) Identify the Test Statistic and its Distribution p= ˆ n +n ...
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