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Multi part question set

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you!! Diversity - 1, MULTIPLE CHOICE S Thank There are fequestions in this set As you proceed through these questions, use necessary information in the previous questions. Recent research shows that diversity is correlated with profitability ("The Business Case for More Diversity". Wall Street Journal, Oct 26, 2019). This problem uses results of that research but makes additional assumptions about sampling and sample values. See the answer key for details. In a sample of 31 firms with the highest diversity ranking, the mean operating profit margin was 12 percentage points, and the standard deviation was 1.4 percentage points. In a sample of 21 firms with the lowest diversity ranking, the mean operating profit margin was 8 percentage points, and the standard deviation was 1.1 percentage points. Assume that operating profit margins were approximately normally distributed in the population of firms with the highest diversity ranking ("high-diversity firms"), and in the population of forms with the lowest diversity ranking ("low-diversity firms”). Assume that the variance of operating profit margins is the same in these two populations, although the value of the common variance is unknown. Assume the samples were independent random samples. Let's begin by testing the null hypothesis that the mean operating profit margin of high-diversity firms is the same as the mean operating profit margin of low-diversity firms, against the alternative hypothesis that the means are different, using a 5% significance level. Which of the following statements describes the basis for the hypothesis test? ONE. X-Y-(x - HY) - N(0,1) by the Central Limit Theorem because we have large random samples from two general (arbitrary) populations with known variances X-Y-W-HY) - N(0.1) by the Central Limit Theorem because TWO. THREE we have large random samples from two general (arbitrary) populations with unknown variances X-Y-xxv) trandom variable, because we have independent random 1 1 Vox my samples from two normal populations with a common but unknown variance X-Y-x-hy) approximate t random variable, because we have FOUR independent random samples from two normal populations with unknown and different variances

Answer 1

1) option 3 is correct , populatiom variance is unknown but equal and two samples are independent so t test will be more appropriate to choose

2) t tablec)

t.50 t.80 t.95 t.975 t.995 t 9995 cum. prob one-tail two-tails 0.50 1.00 t.75 0.25 0.50 0.20 0.40 t.85 0.15 0.30 t.90 0.10 0.20 0.05 0.10 0.025 0.05 t.99 0.01 0.02 0.005 0.01 t .999 0.001 0.0005 0.002 0.001 df 11 2 3 4 5 6 7 8 9 10 11 121 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 80 100 1000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0% 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.678 0.677 0.675 0.674 50% 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.848 0.846 0.845 0.842 0.842 60% 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.045 1.043 1.042 1.037 1.036 70% 3.078 6.314 12.71 1.886 2.920 4.303 1.638 2.353 3.182 1.533 2.132 2.776 1.476 2.015 2.571 1.440 1.943 2.447 1.415 1.895 2.365 1.397 1.860 2.306 1.383 1.833 2.262 1.372 1.812 2.228 1.363 1.796 2.201 1.356 1.782 2.179 1.350 1.771 2.160 1.345 1.761 2.145 1.341 1.753 2.131 1.337 1.746 2.120 1.333 1.740 2.110 1.330 1.734 2.101 1.328 1.729 2.093 1.325 1.725 2.086 1.323 1.721 2.080 1.321 1.717 2.074 1.319 1.714 2.069 1.318 1.711 2.064 1.316 1.708 2.060 1.315 1.706 2.056 1.314 1.703 2.052 1.313 1.701 2.048 1.311 1.699 2.045 1.310 1.697 2.042 1.303 1.684 2.021 1.296 1.671 2.000 1.292 1.664 1.990 1.290 1.660 1.984 1.282 1.646 1.962 1.282 1.645 1.960 80% 90% 95% Confidence Level 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.390 2.374 2.364 2.330 2.326 98% 63.66 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.660 2.639 2.626 2.581 2.576 99% 318.31 636.62 22.327 31.599 10.215 12.924 7.173 8.610 5.893 6.869 5.208 5.959 4.785 5.408 4.501 5.041 4.297 4.781 4.144 4.587 4.025 4.437 3.930 4.318 3.852 4.221 3.787 4.140 3.733 4.073 3.686 4.015 3.646 3.965 3.610 3.922 3.579 3.883 3.552 3.850 3.527 3.819 3.505 3.792 3.485 3.768 3.467 3.745 3.450 3.725 3.435 3.707 3.421 3.690 3.408 3.674 3.396 3.659 3.385 3.646 3.307 3.551 3.232 3.460 3.195 3.416 3.174 3.390 3.098 3.300 3.090 3.291 99.8% 99.9% Z

pooled standard deviation =

s²_p = (SS₁ + SS₂) / (df₁ + df₁)

s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 3.17 / 50 = 1.66

t statistic (under null hypothesis both population means are equal)

t = xbar -y bar / pooled standard deviation *root(1/nx +1/ny)

t = xbar-ybar / 1.66*root((1/31 +1/21))

critical value at 5% level of significance for two tailed test

t critical = +/- 2.008559

if t calculated > t critical , reject null hypothesis ( for right tailed test)

case 1 = tcritical =+ 2.008559

xbar -ybar > 2.008559 * 1.66*root((1/31+1/21))

xbar-ybar > 0.98

cv2 =0.98

case 2 = t critical = -2.008559 (for left tailed test)

xbar-ybar < -2.008559* 1.66*root((1/31+1/21))

xbar-ybar<-0.98

cv1 =-0.98

cv1 =-0.98, cv2=0.98

e) xbar - ybar = 12-8 =4

xbar -ybar >0.98

reject the null hypothesis , xbar-ybar>cv2