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# Statistics for the Business Sciences STAT 133

OSU

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This 8 page Class Notes was uploaded by Alison Vandervort on Monday September 21, 2015. The Class Notes belongs to STAT 133 at Ohio State University taught by Staff in Fall. Since its upload, it has received 62 views. For similar materials see /class/210000/stat-133-ohio-state-university in Statistics at Ohio State University.

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Date Created: 09/21/15

Chapter 9 Key Ideas Hypothesis Test Two Populations Section 91 Overview In Chapter 8 discussion centered around hypothesis tests for the proportion mean and standard deviationvariance of a single population However often researchers want to compare two different populations For example surgeons may want to try out a new surgical technique for a certain ailment but they are not sure if it will be better To test this they could take two samples of people In one sample they could use the existing technique and in the other they could use the new technique By comparing survival proportions of the two groups they could then determine whether the new sample is any better In this case the first population is all people who would receive the existing technique The second population is all people who would receive the new technique y using the methods discussed in this chapter such inference can be done Section 92 Inferences About Two Proportions To use the methods described in this section we first need to rely on a few conditions that must be met for everything to work properly Conditions 1 The proportions are taken from two simple random samples which are independent Here independent means that observations from the first population are not related to or paired with observations from the second population 2 For each of the two samples there are at least 5 successes and 5 failures Notation p1 Population Proportion Population 1 p2 Population Proportion Population 2 n1 Sample Size Population 1 n2 Sample Size Population 2 x1 Number of Successes Population 1 x2 Number of Successes Population 2 31 x71 Sample Proportion Population 1 32 xi Sample Proportion Population 2 n1 n2 il lai fz1 132 7 M Pooled Sample Proportion n1 n a 17 i7 The Test The goal is to test the hypotheses given by H03I71 Pz Hum 172 orpl gt172 orpl ltpz In other words we test whether the proportions for each population are equal vs whether they are different in some way Test Statistic Z 31 7 307071 7 pl 01 Z 131 7 132 For the test statistic we assume Ho 1711 1711 1711 fan is true which means p1 p2 0 n1 n2 n1 quot2 The critical values and PValues come from the standard normal distribution Decisions are made in exactly the same way as in Chapter 8 Example Close to an election ballot issues 3 and 4 are very controversial Researchers want to see whether there is a difference in the proportions of people who support issue 3 and those who support issue 4 They randomly sample 200 total people They ask 100 of these people whether they support issue 3 to which 56 say yes They ask the other 100 people whether they support issue 4 to 05 which 45 say les Is there a difference in the proportions of supporters for each issue Test this With 01 0 Solution From the information given we see that n1 100 x1 56 71 056n2 100x2 45132 045 7 x1 x2 5645 7101 7774505 20495 p n1n2 100100 200 1 H03P1P2 H13P1 P2 Testsm s c Z P1 P2 7 0567045 7 011 7156 g 00707 quot1 quot2 100 100 Traditional Method 0 Since H1 has a i sign we want to nd the critical value that has an area of 02 above it and 012 below it on the standard normal distribution ie we want the value Z 5 2 mural Rtgmn Lnuta Raglan 0 From the table this cutoffvalue with an area of0025 above is 196 0 Now we compare the test statistic to 196 and nd that Z 156 lt 196 o This means that Z is not in the critical region shaded area 0 Therefore we do not reject H0 In other words we conclude that there is not enough evidence to claim that there is a difference in proportions of supporters for the two A 2 9 issues PValue Method 0 Since H1 has a i sign we want to nd area above 121 156 and below 1Z1 156 for the standard normal distribution 0 From the ZTable this area is 200594 01188 0 Now we compare this area to 01 and see that 01188 gt 005 0 Therefore we do not reject H0 In other words we conclude that there is not enough evidence to claim that there is a difference in proportions of supporters for the two issues T 1215 xample Gloria a hairdresser claims that she is better than a fellow hairdresser named Jules They decide to run a hypothesis test to see if this is true Out of 92 ofGloria s customers 87 are satis ed with their haircut Out of 67 of Jules customers 56 are satis ed Run the test to see if Gloria s customers have a higher percentage of satisfaction than Jules customers on 005 Solution From the information given we see that 11192 x187 71 0946n2 612 56 72 0836 x1x2 8756g0899 Fowl P n1n2 9267 159 H03P1P2 H13P1gtP2 TestStatistic Zi FITPZ 7 0946 0836 7 011 7227 EE 0048 quot1 quot2 92 67 Traditional Method 0 Since H1 has a gt sign we want to nd the critical value that has an area of on above it on the standard normal distribution ie we want the value Z a Lnuta Rtgmn 0 From the table this cutoffvalue with an area of005 above is 1645 0 Now we compare the test statistic to 1645 and fmd that Z 227 gt 1645 0 This means that Z is in the critical region shaded area above 1645 0 Therefore we reject H0 and conclude that Gloria has a higher satisfaction percentage an Jules PValue Method 0 Since H1 has a gt sign we want to nd area above Z 227 for the standard normal distribution 0 From the ZTable this area is 1 7 09884 00116 0 Now we compare this area to on and see that 00116 lt 005 94am 0 cm 0 Therefore we reject H0 and conclude that Gloria has a higher satisfaction percentage an Jules Confidence Interval for ELLE Sometimes one would like to estimate the difference between the proportions rather than just seeing whether the proportions differ significantly To construct a confidence interval for 71 7 72 we have the following Point Estimate 31 7 32 Critical Value Z A Standard Error n1 quot2 This gives the following confidence interval 131 41 m3 quot1 quot2 A 7quot i2 P1 172 Example Consider the GloriaJules hairdresser example from the previous page We had the following quantities n1 92 x1 87 31 0946 m 67 x2 56 32 0836 Therefore a 95 confidence interval would be A A 0946 0054 0836 0164 plipziZa 094670836i196 A n1 n2 92 67 0llrl960051 0llr009997 Notice that 0 is not included in this confidence interval Due to this fact we could say that there is a difference between the two proportions ie we would reject H0 in favor ofH1p1 p2 Section 93 Inferences About Two Means Independent Samples In similar fashion to testing for differences in proportions one may also wish to test for a difference in the means of two populations In the interests of time we will consider the most general case where the population standard deviations are unknown and no assumptions are made about them Better hypothesis tests exist when these value are both known or are unknown but assumed to be equal Consult the textbook for more information on these tests Again certain requirements must be met for the techniques discussed in this section to be theoretically sound Conditions 1 61 and 62 are unknown and no assumption is made about the equality of 61 and 02 2 The two samples are independent 3 Both samples are simple random sam les 4 Either both populations are normally distributed or n1 and n are both greater than 30 Notation ul Population Mean Population 1 Hz Population Mean Population 2 n1 Sample Size Population 1 n2 Sample Size Population 2 E1 Sample Mean Population 1 E2 Sample Mean Population 2 s1 Sample Standard Deviation Population 1 s2 Sample Standard Deviation Population 2 Degrees of Freedom df m1nn1 n2 7 1 so df is still n 7 l but here n is the smaller of the two sample sizes The Test The goal is to test the hypotheses given by H03 H1 H2 leui tz 0TH1gtH20139H1 ltH2 In other words we test whether the means for each population are equal vs whether they are different in some way Test Statistic W or 1 E For the test statistic we assume Ho 312 322 312 322 is true which means u1 u 0 n1 n2 n1 quot2 The critical values and PValues come from the Student t distribution with Degrees of Freedom minn1 n2 7 1 Decisions are made in exactly the same way as in Chapter 8 Exampl Researchers want to see if the average money spent per week by tourism in Chicago is less than the average money spent per week by tourists in New York City They take a sample of60 tourists from Chicago and 56 tourists from NYC Ofthe Chicago tourists average spending was 635 with a standard deviation of 50 Of the NYC tourism the average spending was 650 with a standard deviation of30 Run a hypothesis test with a 005 Solution From the information given we see that n160 i1635s150n2 56Z 650s2 30dfminn1n27l 567155 H01 1 H2 H11 1 lt 2 Test Statistic t M i 71974 x12 522 502 302 7599 r11 r12 60 56 Traditional Method 0 Since H1 has a lt sign we want to nd the critical value that has an area ofa below it on the t distribution ie we want the value ta 0 From the table this cutoffvalue with an area of005 below is l 673 M g m 0 Now we compare the test statistic to l673 and nd that t l 974 lt l673 o This means that t is in the critical region shaded area 0 Therefore we reject H0 The evidence suggests that Chicago tourists pay less per week than those in New York City T T 149 4 6393 P Value Method 0 Since H1 has a lt sign we want to nd areabelow t l974 for the t distribution A 0 Now notice that the tTable does not allow one to directly nd this area However for df 55 we see that 2004 has an area below of0025 and l673 has an area below of005 0 Since 2004 lt tlt l 673 the pvalue will fall between 0025 and 005 see picture 0 Now we compare this area to a and see that 0025 ltp lt 005 a o Sincep lt 005 we reject H0 The evidence suggesw that Chicago tourism pay less per week than those in New York City Example One question on everyone s mind is whether there is a difference in the average number of pets owned by Columbus families and the average number of pets owned by Cleveland families Researchers set out to answer this important question They sampled 35 Columbus families and 48 Cleveland families Of the Columbus families the average number of pets was 24 with a standard deviation of 14 Of the Cleveland families the average number of pets was 19 with a standard deviation of 09 Run a hypothesis test on 005 to see ifthere is a difference in the average number ofpets owned by families in the two cities Solution From the information given we see that n1 35 XI 24 s1 14 n2 48Z 19 s2 09 dfminn1n27 1 35 7134 H03 111 H2 H13 H1 i 112 Test Statistic M 1852 ii 142 092 02700 rt1 rtZ 35 48 Traditional Method my nii zdx 0 Since H1 has a i sign we want to nd the critical value that has an area of 02 above it and 02 below it on the t distribution ie we want the value 1 Z nut Regmn 1 1 0 From the table this cutoffvalue with a onetailed area of0025 is 2032 0 Now we compare the test statistic to 2032 and nd that t 1852 lt 2032 0 This means that t is not in the critical region shaded area Therefore we do not reject H0 There is not suf cient evidence to conclude that the average number of pets is different in the two cities P Value Method 0 Since H1 has a i sign the pvalue will be the area above ltl 1852 and below M 1852 for the t distribution 0 Again notice that the tTable does not allow one to directly nd this area However for df 34 we see that 1691 has a twotailed area of010 and 2032 has a twotailed area of 005 0 Since 1691 lt t lt 2032 the pvalue will fall between 005 and 010 see picture 0 Now we compare this area to 01 and see thatp gt 005 01 TTT 1691 m 2031 0 Since p gt 005 we do not reject H0 There is not suf cient evidence to conclude that the average number of pets is different in the two cities Confidence Interval for ELLE As in the previous section one might like to estimate the difference between the means rather than just testing for the difference To construct a confidence interval for M 7 Hz we have the following Point Estimate f1 7 x Critical Value Ia df minn1 n2 7 l 2 Z 2 Standard Error Si S72 quot1 quot2 This gives the following confidence interval 2 Z s s x1 7 x2 i 1 i 72 A n1 n2 Example Consider the avg number of pets example from the previous page We had the following quantities n1 35 E1 24s1 14 ml 48 E2 19 Si 09 dfminn1 n27 1 35 71 34 Therefore a 95 confidence interval would be 142 092 35 48 05203202700 05i05485 2 4 S2 S2 Ewan ii24719i2o32 4 n1 n2 Notice that 0 is included in this confidence interval Due to this fact we could say that there is not a significant difference between the two means ie we do not reject H0 in favor ole ul uz as in the example on the previous page

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