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NEED HELP ASAP - will give thumbs up Multi part question set If you can’t finish...

Question

NEED HELP ASAP - will give thumbs up Multi part question set If you can’t finish...
NEED HELP ASAP - will give thumbs up
Multi part question set
If you can’t finish all of it, that’s ok!
Would really appreciate any and all help as soon as possible
Thank you so much in advance!

you!! Diversity - 1, MULTIPLE CHOICE S Thank There are fequestions in this set As you proceed through these questions, use ne
ONE TWO FOUR THREE Give the table value (value of z, or ty) that you will use when you calculate the critical values in the n
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
ONE TWO FOUR THREE Give the table value (value of z, or ty) that you will use when you calculate the critical values in the next two questions. (zu or tu is a positive value.) Four decimals 2. The form of the rejection region is: Reject Ho if X-Y<cv, or if X-Y>cvz. The "X" population is high- diversity firms and the "Y" population is low-diversity firms. Calculate the value of cv1. Two decimals 3 Calculate the value of cv2. Two decimals 4 Which of the following accurately describes the conclusion of the hypothesis test and the reason for it? ONE. Accept Ho because cv> <ă-Y<cv2 TWO. Reject H, because X-Y<cva THREE. Reject Hy because X-F>CV2 ONE 5 THREE TWO

Answers

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 tablet.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.c)

pooled standard deviation =

s2p = (SS1 + SS2) / (df1 + df1)

s2p = ((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


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