427 0092 519 519 there is a 95 chance that true

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There is a 95% chance that true proportion of cheaters in college of business at Bayview University falls between 33.5% and 51.9%.Proportion of Male Cheatersp= 27/47n = 47p= .574z = 1.96p = .574______________.574 ± 1.96(.574(1-.574))/47.574 ± .141Lower Limit: .574 – 0.141 = .433 = 43.3%Upper Limit: .574 + 0.141 = .715 = 71.5%There is a 95% chance that true proportion of male cheaters in college of business at Bayview University falls between 43.3% and 71.5%.Proportion of Female Cheatersp = 20/47n = 47p = .425z = 1.96
ETHICAL BEHAVIOR OF STUDENTS AT BAYVIEW UNIVERSITY6 p = .425______________.425 ± 1.96(.425(1-.425))/47.425 ± .141Lower Limit: .425 – 0.141 = .284 = 28.4%Upper Limit: .425 + 0.141 = .566 = 56.6%There is a 95% chance proportion of female cheaters in college of business at Bayview University falls between 28.4% and 56.6%.Mean GPA of Cheaters and Non CheatersI did a two-sample test to compare the number of cheaters and the non-cheaters GPA.The survery shows that the average GPA of a cheater is 3.25 and the average GPA for non-cheaters is 2.98. These results suggest that the GPA for business students who cheat is better thanthe GPA for business students who do not cheat. Here I have set up directional hypothesis testbecause the GPA for cheaters is higher. My alpha is .05 and the t-critical value for one tailed testis 1.67. GPA CheatersXC= 3.25SC= .5nC= 47GPA Non-CheatersXN= 2.98SN= .49nN= 63
ETHICAL BEHAVIOR OF STUDENTS AT BAYVIEW UNIVERSITY7 H0: μC= μNHA: μCμN1-tailed test=.05αn = 47 + 63 = 110Degree of freedom =110 - 2 = 108tcrit= (XC– XN– (μCμN))/ SxC– xNspooled= ((nC-1) s2C+ (nN-1) s2N)/ nC+ nN– 2= ((47 – 1) .5^2 + (63 - 1) .49^2)/ 47 + 63 -2= 26.386/108= .244______________________sx-bar for C – x-bar for N= ( s2pooled/ nC) + (spooled/ nN)___________= .0051 + .0038_____= .0090sx-bar for C – x-bar for N= .09518tcalc= (x-bar for C – x-bar for N– (μT- μD)/ sx-bar for C – x-bar for N= (3.25 – 2.98)/ .09518= (.27) / .0951= 2.83Therefore I rejected the H0because my t-calc is larger than my t-critical value.
ETHICAL BEHAVIOR OF STUDENTS AT BAYVIEW UNIVERSITY8 Male vs. Female CheatersIn this section I am testing a hypothesis to test if male students cheat more than femalestudents. My alpha is .05 which therefore my z-critical value is 1.65 absolute. I plugged the datain the formula and calculated my z-calc. H0: πM= πFpM= 27/60 = .45HA: πM> πFpW= 20/50 = .40=.05αZcrit= |1.65|Men CheatersnM= 60pM= .45PC= 47/110 = .427Women CheatersnF= 50pF= .40PC= 47/110 = .427______________________________SpM– pF= (.427(1-.427))/60 + (.427(1-.427))/50____________= (.0041+.0049)____= .009= 0.095Zcalc= (p1 – p2)/ (SpM– pF)= (.45-.40)/.095= .5263
ETHICAL BEHAVIOR OF STUDENTS AT BAYVIEW UNIVERSITY9 - .5263 < 1.65 (Zcrit) I fail to reject the null hypothesis. There is no statistical evidence to prove that male

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