Educational Statistics EDUR 8131
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Date Created: 10/12/15
Notes 5 t tests 1 The one sample ttest a Formulas Recall the Z test formula Z Q X UJ The one sample ttest which is very similar to the Z test has the following formula 21141 where the only difference is SD versus 6 in the Z test That is the standard error of the mean is now estimated by the formula Q SX 5 where the symbol s is used to indicate that the standard error of the mean is being estimated with the sample SD Recall that the standard error of the mean for the Z test was calculated as b Hypotheses for the one sample ttest Hypotheses for the one sample ttest are formulated in exactly the same manner as in the one sample Z test Using SAT as an example note u 1000 the one sample ttest nondirectional hypothesis is symbolized as Nondirectional H1 p 1000 HO 1 1000 Directional onetail tests 1 Lowertailed test H1 1 lt 1000 HO 1 2 1000 or HO 1 1000 this one is preferred Uppertailed test H1 1 gt 1000 HO 1 g 1000 or HO 1 1000 this one is preferred Note that for the directional hypotheses the alternative H1 states what one expects to nd as long as a relationship or difference is expectedh For example if one expects that a sample of students will have a higher than average IQ then H1 1 gt 1000 Similarly if one expects that a given sample of students will have a lower than average IQ the H1 1 lt 1000 c Critical tvalues tent Like the Z test one may use critical values for hypothesis testing Critical tvalues are obtained from a t table see text Note that tvalues have distributions that are similar to normal distributions but they are slightly fatter in the tails Finding tvalues in the t table is similar to the Z table To nd the correct critical tvalues denoted as tent one must rst calculate the degrees of freedom df or v For the one sample ttest degrees of freedom are de ned as dforvn1 where as before 11 is the sample size Degrees of freedom can be described as the amount of information available in the sample after certain mathematical restrictions are applied to the data d Statistical signi cance on Next one must determine the level of statistical signi cance for the analysis As before alpha is usually set at 10 05 or 01 Once the alpha level is determined critical values for the one sample ttest can be found e Decision rules Deciding whether to reject or fail to reject the null can be determined by decision rules The decision rules are Twotailed tests Ift g tmt or t 2 tent then reject H0 otherwise fail to reject H0 Onetailed uppertailed test Ift 2 tent then reject H0 otherwise fail to reject H0 Onetailed lowertailed test Ift g tent then reject H0 otherwise fail to reject H0 Note that t t symbolizes the critical tvalue found in t tables and is different from t which is the calculated tratio obtained from sample data f An example A physical education teacher wishes to know whether his class of students is statistically above or below the national average in weight The national average for eighth graders is p 100 The student weights for his class are 99 98 105 110 115 103 88 125 130 and 115 For this sample n 10 2 1088 SD 12839 so Yip Y71088100 88 88 55 5 1283mm 128393162 406 39 Version 2142005 The dforvn 1 10 1 9andsetoc 05 The goal of the test is to determine whether the sample has an average weight that is statistically different from the national average This calls for a nondirectional test because the specific direction of the mean difference higher or lower was not indicated Therefore HO 1 100 and H1 1 i 100 The critical value is tmt 2262 The decision rule is If2167 g 2262 or 2167 2 2262 then reject H0 otherwise FTR H0 Since 2167 is neither less than 2262 nor greater than 2262 the null is not rejected ie fail to reject and one concludes that the sample does not have a statistically different mean from the national average What happens if instead it was hypothesized that the sample would have a lower than average weight ie H1 1 lt 100 and H0 1 100 This is a lower tailed test since the sample is expected to have a lower mean score If on 05 the corresponding critical tvalue is tmt 1833 The decision rule states If2167 g 1833 then reject H0 otherwise fail to reject H0 Again the null is not rejected ie fail to reject Finally had one hypothesized that the sample would be above average then H1 ugt 100 and HO 1 100 This is an uppertailed test The critical tvalue for an alpha of 05 is ted 1833 and this time the null is rejected so the sample can be said to have a statistically higher average weight than normal It is also possible to perform hypothesis testing with the ttest without using critical tvalues Recall that the Z test had decision rules for pvalues Calculated probability values pvalues are usually reported with statistical software and the decision rules for the Z test also apply to the t test See earlier notes for these decision rules This section to be discussed further in class g Assumptions The assumptions for the one sample ttest are identical to the Z test normality and independence Version 2142005 h Exercises 1 Raw MAT scores are 31 38 27 41 39 and 36 Is this sample statistically different from the national average MAT of 30 Set on 01 2 Same scores but on 05 3 Same scores and 0c 05 but this time hypothesize that the sample average will be greater than population average 4 Fifteen students have a sample MAT mean of 323 with a sample standard deviation of 473 Does this sample of students have a mean MAT score that is statistically different from the national average at the 01 level with a two tailed test What about on 10 and a two tailed test What about on 05 and an uppertailed test 5 Suppose the following random sample of ITBSmath scores are observed in your middle school 45 58 65 63 35 43 78 55 58 69 81 and 49 Is this evidence that your middle school has a student population above average in terms of mathematics skills national average ITBSmath is 50 Set alpha at 6 You wish to determine whether you are getting cheated every time you buy a bag of apples The standard bag of apples that you buy states that it contains one pound 16 ounces of apples After you get home you notice that the bag only contains 155 ounces not the stated 16 ounces To determine whether or not the company is systematically cheating the consumer you decide to buy every 16 ounce bag of apples in the three local grocery stores After weighing each bag you nd the following weights 143 155 163 170 152 159 148 150 152 159 157 156 and 161 Setting the signi cance level at 05 does it seem the company is systematically cheating the consumer Which should you perform an upper lower or twotailed test Why 7 Ford claims that its new car Aspire gets 39 mpg on the highway Consumer Reports magazine wishes to test this claim so they hire you for 1500 to perform the statistical testing They buy 10 Aspires and road test each They nd the following mpg estimates for the cars 32 43 39 38 34 36 35 38 39 and 36 Their question to you is Does our sample of Aspires have an estimated mpg that is different from Ford39s claim Set alpha at 05 and give them an answer Computer output for exercises 1 through 7 Exercises 1 through 3 ttest mat3O Variable l Obs Mean Std Dev mat 6 3533333 5316641 Ho mean 3O t246 withS df Pr gt ltl 00574 Version 2142005 Exercise 4 ttesti 15 323 473 30 Variable 1 Obs Mean Std Dev x 1 15 323 473 Ho mean 30 t 188 with 14 df Pr gt 1t1 00806 Exercise 5 test itbs5O Variable 1 Obs Mean Std Dev 1TBS 1 12 5825 1393573 Ho mean 50 t 205 with 11 df Pr gt 1t1 00649 Exercise 6 ttest weight 16 Variable 1 Obs Mean Std Dev weight 1 13 1557692 7013722 Ho mean 16 217 with 12 df Prgt1t100503 Exercise7 ttest mpg 39 Variable 1 Obs Mean Std Dev mpg 1 10 37 3091206 Ho mean 39 t 205 with 9 df Pr gt 1t1 00711 2 Con dence Intervals CI for Means When estimating a parameter one typically uses a point estimate like 2 s or s2 Using these point estimates one may construct an interval which will show a possible interval range of values which might include the parameter being estimated A con dence interval CI for p is found by 139 OOCI Y i leathf S which stated differently is Version 2142005 1ocCI1 t s Ira2 Critical which is 7 SD 139 OOCI X i paztmnmz or simply 7 SD 7 SD 1 39 00C1 X naztummz jaXJrnaztcnncazK D This is 1001 0c con dence interval That is ifoc 05 then this is a 1001 05 10095 95 con dence interval or 95CI A 95CI means that one can be 95 con dent that all intervals constructed like this for 100 random samples in the long run will contain the population value 1 This means that if 100 such intervals were constructed on average the population value of 1 would be correctly included in 95 of those intervals while would increase fail to include 1 To calculate this CI choose on say at 05 then construct the interval by simply nding the critical value associated with on 05 and lling in the rest of the formula Example Construct 95CI for a class of high school students 11 12 with a mean IQ of 120 and a standard deviation of 165 120 7 2201 120 2201 M E 1207 2201 x 4763120 2201 x 4763 120404831204104113 109517 130483 With such an interval one may state that one is 95 con dent that this interval contains the true 1 for all students who are like the students in the particular high school class apparently smart students Based upon this con dence interval it seems that this high school class is quite different from the mean score typically found for IQ tests in the population How does one know this The CI may also be used as a nondirectional hypothesis test If the hypothesized population value of p is not within the CI then H0 1 100 may be rejected Since the value 100 is not within the interval constructed which ranges from 1095 to 1305 one may conclude that sample data appears to differ statistically from the hypothesized value of 100 In this particular case the sample data such a mean that is higher than the expected value of 100 Version 2142005 Exercises 1 Construct a 99CI for the following scores 120 123 125 101 98 101 Test the hypothesis using the 99CI that Ho 11 100 2 Same as 1 but use a 95CI 3 Same as 1 but use a 90CI 4 Fifteen students have an SAT mean of 1200 with a standard deviation of 150 Does this sample of students have a mean SAT score statistically different from the population mean of 1000 at the on 05 Use a CI to answer this question Is the mean statistically different if a 99CI is used 3 The TwoIndependent Samples ttest also called the Two Group t test a Situation Both the Z test and the one sample ttest allow one to statistically comparing the mean of one sample of observations with a given population value eg u If one is interested in comparing two independent groups then the two independent sample ttest may be appropriate For example suppose one is using a posttest only control group design to examine the effect of computer assisted learning in geography achievement among third graders The control or comparison group is taught US geography with the traditional methods using maps textbooks and workbooks The experimental group uses the computer game Where in the US is Carmen SanDiego At the end of the lesson both groups are given the same posttest A two group independent ttest would be appropriate for determining statistical difference between the control and experimental groups b Hypothesis formulation One may formulate three different research hypotheses for the above example Nondirectional The experimental and control group will have different levels of achievement in US geography Ho M1 M2 and H15 M1 i M2 or HO 111 112 000 and H1 111 112 i 000 where 111 represents for group 1 experimental group and 112 represents group 2 control group Directional group 1 has higher mean than group 21 The experimental group will show a higher level of achievement Ho M1 3 M2 and H15 M1 gt M2 or HO 111 112 g 000 and H1 111 112 gt 000 Version 2142005 Directional group 2 has higher mean than group 11 The experimental group will show a lower level of achievement Ho M1 2 M2 and H15 M1 lt H2 or Hop1 Hz 2 000 and H1p1 uz lt 000 c Formulas for calculating the t ratio To test the above hypotheses the two sample independent t statistic is calculated as t Y1 Yz 11 712 San Since it is usually assumed that M uz 000 no difference in the population values the t formula can be simplified to t X1 X2 X1 X2 sxrgz SE0 where 52 52 magi n1 quot72 2 Note that SEd represents the standard error of the difference which like the standard error of the mean represents the standard deviation of the sampling distribution for X1 7X2 The symbols sf and 322 represent the variances for group 1 and group 2 respectively Recall that the sampling distribution of the sample mean has a known distribution that approaches the normal distribution when sample sizes are large The sampling distribution for Y1 7X2 also follows the central limit theorem Note that the mean of the sampling distribution of X1 7 Y2 is equal to M Hz The standard error for Y1 7X2 is SEd s272 d Degrees of Freedom Degrees of freedom for the two independent sample ttest are dforvn1n272 where the 111 is the sample size for group 1 experimental group and n2 is the sample size for group 2 control group Version 2142005 e Decision Rules The decision rules are the same as for the one sample ttest Twotailed test Ift g tmt or t 2 tent then reject H0 otherwise fail to reject H0 Onetailed uppertailed group 1 anticipated to have higher mean than group 2 test Ift 2 tent then reject H0 otherwise fail to reject H0 Onetailed lowertailed group 1 anticipated to have lower mean than group 2 test Ift g tent then reject H0 otherwise fail to reject H0 f Assumptions The two independent samples ttest requires that the raw scores in both populations be normally distributed and independent Also the two populations should have equal homogeneous variances The two group ttest is generally robust to nonnormality and unequal variance provided n1 m n2 but is not robust to dependence of observations g An Example Recall the geography experiment The scores for both groups are Experimental Group Control Group 88 79 89 75 91 86 95 91 86 92 87 82 88 80 79 82 88 81 72 87889 75 83111 s 4256 s 5578 n 9 n 9 The experimental group has a mean of 87889 and a standard deviation of 4256 and the control group had a mean of 83111 and a standard deviation of 5578 There were 9 students in the experimental group and 9 students in the control group So the two independent group ttest with an on 05 and a non directional test would be t 2ch 4778 2043 Sg 3 18114 31114 2339 9 9 and the degrees of freedom are df n1 n2 2 9 9 2 16 The critical t is ted 2120 The rejection regions are t f 2 120 and t E 2120 and the decision rule is Version 2142005 If2043 g 2120 or 2043 2 2120 then reject H0 otherwise FTR H0 The correct decision is fail to reject HO One would therefore conclude the following There is not a statistically significant difference in geography achievement between the experimental and control group for this sample at the 05 level of signi cance This finding indicates achievement scores for geography students do not appear to differ between those who do and do not use the software Carmen SanDiego Note however what would happen if one hypothesized that the experimental group would have higher scores than the control group If on 05 the critical value for an uppertailed would be 1746 so the decision rule would be If2043 2 1746 then reject H0 otherwise fail to reject H0 Now HO is rejected and one could conclude the following The data indicate that students who learn with the computer program Carmen SanDiego show a statistically signi cant at the 05 level higher achievement score in US geography Thus use of the software appears to benefit students h Con dence Intervals About Mean Differences Recall the CI for a sample mean 1ocCIi t s Ira2 Critical One may similarly compute a CI for the difference between two means The formula is locCITli2r t sgrgz Ira2 Critical The 95CI for the above example is 3995CI Y1 7Y2 i 975t57121501S 1 2 87889 83111r 2122339 4778 r 4959 or between 0 181 and 9737 Since 0 is within this interval HO will not be rejected Version 2142005 Computer Analysis of Above Example ttest scores bygroup Variable l Obs Mean Std Dev 0 l 9 8311111 5577734 1 l 9 8788889 4255715 combined 1 18 855 5404247 Ho meanx meany assuming equal variances t 204 with 16 d f Pr gt ltl 00579 i Strength of Association for Two Group ttest effect size While a statistically significant ttest indicates that the two groups are probably not equal the ttest does not indicate the strength of the association between the independent variable and the dependent variable In the study just discussed the independent variable IV is the presence or absence of the treatment and the dependent variable DV is the posttest achievement score The question one may ask after rejecting H0 is just how strong an impact does the treatment have on student achievement One measure of the strength of the association between the treatment and the outcome is eta squared 112 2 t2 n 1 t2 df For example the calculated t above was 2043 so 2 1127 2043 4174 207 2043 16 417416 The value obtained for 112 may be interpreted in a manner identical to 12 such as the variance explained or predicted in posttest scores by the treatment In fact if one calculates a Pearson39s correlation between the two numerical variables listed in the table below posttest scores and the indicator of treatment ltreatment 0control the obtained r will be equal to 455 and the r2 will be 207 Version 2142005 Posttest Scores Indicator of Treatment Treatment Condition 8 8 1 Experimental 89 1 Experimental 91 1 Experimental 95 1 Experimental 86 1 Experimental 87 1 Experimental 8 8 1 Experimental 79 1 Experimental 8 8 1 Experimental 79 0 Control 75 0 Control 86 0 Control 91 0 Control 92 0 Control 82 0 Control 80 0 Control 82 0 Control 81 0 Control This should indicate to you that one may actually use a Pearson correlation to determine whether two groups are statistically different For example using the same experimental data one could reproduce the same t value obtained from the two independent groups ttest using only the correlation r rxn72 45541872 182 7 2043 4142 17207 891 In short the two group independent ttest and the Pearson correlation coef cient provide identical inferential results The two group ttest requires the calculation of 112 in order to determine the strength of the relationship between the IV and DV 139 Effect Size ESQ One may choose to relate to the reader the magnitude of the effect of the treatment by providing 112 Another means of relaying this information which is growing in importance in research today is the standardized ES indicator ES denoted in the researcher literature as d andor A may be calculated with one of two formulas First d is d X1 7 X 2 SDWithin where Version 2142005 SD 2X7Y12ZX7Y22 within r1 1 n2 71 SDWMn is essentially the average SD for the two groups Second A is 2 72 SD A control group where SDcoml gm is simply the SD of the control group if one is present Note that both d and A describe the magnitude of the difference between the two group means in standard deviation units So for example if d or A 2 then this indicates that the two group means differ by 2 standard deviations The larger either d or A the greater the difference between two groups and hence the larger the effect of the treatment In the example used above the ES is 2X7Y1zZX7XZZ quot2 1 144908248913 7D7D 393821 16 2461 4961 2272 8788978311 SD 4961 SDwithin d 0963 within If one wished to calculate A then the corresponding ES is A Yriyz SD 8788978111 5578 0857 control group Either ES is appropriate to use when an experimental group is compared to a control group When two groups are compared and the two groups do not represent experimental and control such as males vs females then one should use d as the measure of ES Version 2142005 k Exercises 1 Determine whether boys have a statistically different at the 1 level ITBS math score from girls The mean math score for boys is 78 s 53 and the mean for girls is 73 s 61 There are 25 boys and 25 girls a What is the correct HO and H1 in both written and symbol form b What are the critical and calculated tvalues 2 Determine whether a statistical difference exists between men and women in weight Men 156158 175 203 252 195 Women 149 119 168 123 155 126 a Test for a nondirectional HO with on 01 what is the correct HO H1 b Test for a nondirectional HO with on 10 c Test the hypothesis that men will have lower weight and set on 10 What is the correct HO H1 3 Two classes of educational research were taught with two different methods of instruction teacher guided TG and self paced SP Which had the better student achievement at the end of the quarter TG scores 95 93 87 88 82 92 SP scores 78 89 83 90 78 86 a Test for a nondirectional HO with on 01 what is the correct HO H1 b Test for a nondirectional HO with on 10 c Test the hypothesis that TG will have higher scores and set on 05 What is the correct HO H1 1 Computer answers to exercises Example 1 ttesti 25 78 53 25 73 61 Variable l Obs Mean Std Dev x l 25 78 53 y l 25 73 61 combined 1 50 755 6193644 Ho meanx meany assuming equal variances t 309 with 48 df Pr gt ltl 00033 Version 2142005 Example 2 ttest weight bysex Variable Obs Mean Std Dev 0 l 6 14 2007984 1 l 6 1898333 3589661 combined 1 12 1649167 3802979 Ho meanx meany assuming equal variances t 297 with 10 df Pr gt t 00141 Example 3 ttest scores bygroups Variable Obs Mean Std Dev 0 l 6 84 525357 1 l 6 895 4764452 combined 1 12 8675 557796 Ho meanx meany assuming equal variances 190 with 10 df Pr gt t 00867 4 Two Correlated Group t test also called dependent samples ttestL The correlated t test allows the researcher to consider differences between two groups or sets of scores that are related to oneanother Under what conditions is one likely to nd correlated or dependent samples or groups Condition 1 BeforeAfter Studies Multiple Measures on the Same Subject This type of data occurs most o en with pretesttreatmentposttest experimental designs These types of designs are used to determine whether some treatment will change posttest scores relative to the pretest score The pretest and posttest scores are related because the scores are taken from the same individuals ie each person is measured twice Examples a A student takes the SAT enrolls in an SAT enhancement class and then retakes the SAT Two scores from the same student exist b A teacher measured the reading performance of a thirdgrader presented some treatment designed to increase reading performance then remeasured the student s reading performance again two scores from same individual c A PE teacher measures the vertical jumping ability of his class provides his class a weight training program for one month then remeasures vertical jumping ability of each student two scores from same students Version 2142005 Condition 2 MatchedSubjects Two groups are involved in the study experimental and control and they are matched on some extraneous variables that is likely to be related to the dependent variable being examined Examples a A teacher is interested in determining whether quotHooked on Phonicsquot increases thirdgrade students39 reading performance Using two groups of students group A the experimental group will use quotHook on Phonicsquot for one month and group B the control will be exposed to the usual reading lessons during the month The teacher knows that IQ in uences reading performance so to control for the effects of IQ on the dependent variable which is a posttest on reading performance the researcher matches students in the two groups on their IQ levels in a fashion similar to the schematic below I GroupAtreatment I GroupBcontrol I IQ score IQ score High 110 Beth and Sue John and Ann High 110 Middle 90110 Bob and Susan Fred and Bill Middle 90110 Low lt90 Bryan and Bill Josh and Walt Low lt90 In this scheme students from both groups are matched according to their IQ levels It is important to match on IQ since we would expect students with higher IQs to perform better on a reading test than students with lower IQs b As another example one might make a comparison of faculty salary between men and women to determine whether sexual discrimination exists It would be important to match men and women on academic rank since we know that assistant professors on average make less than associate and full professors Condition 3 Naturally occurring pairs Natural pairs such as husbands and wives twins brothers sisters brothers and sisters parents and their children etc With naturally occurring pairs one would expect the pairs to hold similar feelings beliefs attitudes etc so their scores will generally be related to oneanother Examples a Determining whether husbands39 attitudes toward politics are similar to their wives Since people tend to marry others like themselves one would expect that most husbands and wives to hold similar political views b Determining whether boys39 IQ differs from girls39 IQ Since brothers and sisters are similar genetically one might anticipate the two to have similar IQs that is their IQs are likely to be related therefore brothers and sisters need to be matched Hypothesis Formulation The hypothesis tested with the correlated ttest is the same as in the independent ttest For example suppose one is in determining whether boys or girls get higher math scores on the ITBS Clearly intelligence plays an important part in determining mathematics performance so this is a factor that needs to be controlled through matching One may formulate several hypotheses as demonstrated below Version 2142005 Nondirectional The average ITBS math scores will differ between boys and girls their scores will differ on average Ho H1 H2 and H15 H1 i H2 or Hou1 uz 000 and H1u1 pi i 000 where M represents for group 1 boys and pi represents group 2 girls Directional group 1 has higher mean than group 2 1 Boys will score higher on average than girls Ho H1 3 H2 and H15 H1 gt H2 or Hou1 uz g 000 and H1u1 uz gt 000 Directional group 1 has lower mean than group 21 Boys will score lower on average than girls Hou1 2 pi and H1u1 lt Hz or Hou1 pl 2 000 and H1u1 uz lt 000 Theoretical Formula for Correlated t test The t ratio for the correlated t test can be calculated as t 2271 527 where fl 7 Y2 is the difference between the two sample means and the denominator is the standard error of the difference SEd Note that this is identical to the formula for the two independent sample t test The difference between the formulas for the independent and the correlated t test occurs in the calculation of the standard error of the difference For the correlated t test the standard error of the difference is calculated as but in the independent t test it is assumed that the groups are not related scores between groups are not correlated so the standard error looses the correlated term in the formula ie Version 2142005 Z Z Z Z Z Z Z Z 7 S1 SZ S1 SZ 7 S1 SZ S1 SZ 7 S1 SZ 7 S1 SZ 7 7 7 7772 7 7 7 77720 7 7 7 77707 77 er n llesz er n2 ful l er n m n If there is no correlation then the SE formula reduces to the SE formula given in the independent samples ttest In short the primary difference between the two t tests is the calculation of the standard error of the difference SEd Practical Formula for Correlated t test To calculate the correlated t statistic the following formula is easier to use Hamid 3 1 sjn sjn SE01 where a is the mean of the differences between pairs of scores ie a 7726 n and SE is the standard error of the differences SEd sjn where s is the variance of the difference scores and is calculated like a regular variance ie 201767 2 S d n71 In short the correlated t test is may be viewed as the mean of the difference d divided by the standard error of the difference SEd t as d Degrees of Freedom The df for the correlated t test is calculated as df n l where n represents the number of pairs across the two groups Decision Rules The decision rules are the same as in the independent twosample t test Version 2142005 Twotailed tests Ift g tmt or t 2 tent then reject H0 otherwise fail to reject H0 Onetailed uppertailed test Ift 2 tent then reject H0 otherwise fail to reject H0 Onetailed lowertailed test Ift g tent then reject H0 otherwise fail to reject H0 Note that t t symbolizes the critical tvalue found in t tables and is different from t which is the calculated tratio obtained from sample data Example 1 Suppose we are interested in determining whether salary differs between men and women faculty at GSU When randomly selecting subjects for the study it is important that we take into consideration their academic rank since full professors make more money than associate professors and associates make more money than assistant professors on average Test the hypothesis of no difference between men and women HO ul pa at the 5 signi cance level Income Difference Income Rank Men Women Rank Full Bill 48000 3000 Beth 51000 Full Full Bob 51000 6000 Bertha 45000 Full Associate Billy 43000 1000 Bobby 44000 Associate Associate Burt 38500 2500 Bonnie 36000 Associate Assistant Brando 24500 500 Brenda 25000 Assistant Assistant Bart S 28000 5000 Bette 23000 Assistant Assistant Brent 33000 7000 Beulah 26000 Assistant 7 d d 27 700 2285714 n Difference Mean of Difference Deviation Deviation Squared D J d 7 a d 7 W 3000 2285714 5285714 2793877249 6000 2285714 3714286 1379592049 1000 2285714 3285714 1079591649 2500 2285714 214286 4591849 500 2285714 2785714 776020249 5000 2285714 2714286 736734849 7000 2285714 4714286 2222449249 SEd sjn where s is the variance of the difference scores and is calculated like a regular variance ie Version 2142005 8992857143 771 7 89928657143j7 1498839524 1463269 so the t value will be t d L22857141562 sdzn SE 1463269 The critical values at the 05 level for df n 1 6 are r 2447 so fail to reject H0 and conclude that salaries do not appear to differ between men and women faculty at GSU even after controlling for academic rank What do you think would happen if an independent samples ttest were used to analyze the above data Calculate the regular independent t test and see Mmen 38000 SDmen 1001249 Mwomen 3571429 and SDwomen 112504 Which is more powerful recall that power represents the probability of rejecting a false HO the independent or correlated t test Why Version 2142005 Example 2 A researcher wishes to discover whether or not the intake of orange juice increases the potassium level in the bloodstream A group of 12 elderly patients are selected from those in a nursing home where previous diet has been controlled Potassium blood levels are measured for each subject Next each subject is given a quart of orange juice and two hours later potassium levels are again measured Test the difference in potassium levels at the 5 level The data are as follows the scaled scores represent potassium blood levels Subject Before After Difference Mean of Deviation Deviation Potassium Potassium Difference 7 Squared Level Level 7 d T d 7 2 d d 7 d 1 26 25 1 2 3 9 2 25 28 3 2 1 1 3 24 27 3 2 1 1 4 23 26 3 2 1 1 5 23 25 2 2 0 0 6 21 23 2 2 0 0 7 19 21 2 2 0 0 8 17 19 2 2 0 0 9 17 16 1 2 3 9 10 16 19 3 2 1 1 1 1 15 18 3 2 1 1 12 14 17 3 2 1 1 7 d 7 d Z 724 200 n 12 and the standard error of the difference is 72 2014 24 24 quot1 1271 H 2182 SE2 7 7 739 426 d S n n 12 12 12 The hypothesis was that orange juice will increase potassium in the blood stream ie the pretest scores will be lower than the posttest scores This hypothesis indicates that a lowertailed test is needed since Ho M1 2 M2 and H1411 lt H2 The critical value at the 05 level for df 12 1 11 is 1796 so we reject HO and conclude that orange juice does appear to increase the amount of potassium in the blood stream for elderly people Version 2142005 Exercises 1 A researcher is interested in determining whether typing speed is affected by the kind of typewriter electric versus manual used A group of student typists equally experienced on both types of machines are randomly selected and are matched on the basis of their typing speed errorfree words per minute One group is then tested on an electric machine and the other group on a manual machine Test HO at the 1 signi cance level The data are as follows a What are the correct HO and H1 in both written and symbolic form b What is are the critical values c What is the obtained calculated tvalue d Did you reject or fail to reject HO e Write your conclusion as if explaining the results to nonstatisticians Pair Typing Speed Electric Manual 1 High 50 42 2 High 65 60 3 Middle 72 65 4 Middle 90 85 5 Middle 48 50 6 Low 62 60 7 Low 75 60 8 Low 50 51 9 Low 68 59 Version 2142005
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