Assume that both pepuletions are nomaly dstrbuted a)Test whether 12p2 athequ0.05

Question

Assume that both pepuletions are nomaly dstrbuted a)Test whether 12p2 athequ0.05 lavel of sentarco k r the gven sarow data b) Construct 95% confidence neral about h" Student t-Distribution Table EEE Cick the icon to viow the Student t-distribution tabie a) Perfom a hypothes's test Determine the null and atemative hypotheses Determine the test stetietic Roure so twe decimal places as neoded) Determine the critical value(s). Select the comect choice below and fil in the answer boxles) within your chaice. Round to tree decimal plapes as needed.) OA The criticeal value i n3s 2 3.4K? 4541 5841 >45i l®.tis L24 07270.9091456 L476 2*15 2.571 2.257 3654.032 4.173 5849 6 9 0,711 708 683 110013 0.es7 0.526 1.088 136, 17% 2.301 2.328 2718 310, Mur 4M5 4437 0,SA 0.520 1.079 L359 1721 2,160 B. The lower critical velae is The upoer crifical vatue is Should the hypothenis be rejected | the hypothesis because the lest saistc 13331 Cick to select your anewers) b) Construct 0 95% confidence interval about -P2. The confidenco ntorval is tne range from (Raund to two deaimal plaoes as needed. Use asconding ordor) Clck to select your answers.

Explanation / Answer

(a) Here the hypothesises are

H0 : 1 = 2

Ha : 1 > 2

Here pooled standard deviation

sp = sqrt [ {(n1 -1)s1 2 + (n2 -1)s22 }/ (n1 + n2 -2)] = sqrt [ (23 * 6.12 + 17 * 11.42 )/ (23 + 17)] = 8.754

Test statistic

t = (x1 - x2)/ sp * sqrt(1/n1 + 1/n2) = (49.8 -41.9)/ [ 8.754 * sqrt (1/24 + 1/18)] = 7.9/ 2.7295 = 2.89

critical value t for dF = 40 and alpha = 0.05

tcritical = 1.6838

so here t > tcritical

so we shall reject the null hypothesis.

(b) 95% confidence interval = (x1 - x2) +- t40,0.05 se0

= (49.8 -41.9) +- 2.021 * 8.754 * sqrt (1/24 + 1/18)

= 7.9 +- 5.5164

= (2.38, 13.42)

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