of North Fon xMclian ill Connect De x D Qiz on Chapter 10Pandor C | ezto.mheduca
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of North Fon xMclian ill Connect De x D Qiz on Chapter 10Pandor C | ezto.mheducation.com/hm.tpx McGraw Hill Connect De Quiz on Chapter 10 Pandora Radio Consider the olowing competing typotheses and accompanying sample data drawn independenty from normally distributed populations. Use Table 2 HOPI-p2 20 1-255 $1-33 12 = 264 82 15 n2-9 a-1. Calculate the value of the test statistic under the assumption that the population variances are unknown but equal. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) est statistic 2001 a2. Calculate the critical value at the 10% level of significance (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) Critical value a-3. Do you reject the null hypothesis at the 10% level? O Yes. since the value of the test statistic is not less than the critical value No, since the value of the test statistic is less than the critical value No. since the value of the test statistic is not less than the critical value Yes, since the value of the test statistic is less than the critical value. b.1. Caliculate the value of the teat staisaic under the assumption that the population variances are Type here to searchExplanation / Answer
Here population variance are equal it is given
(a) pooled standard deviation sp = sqrt [(n1 -1)s12 + (n2 -1)s22 }/(n1 + n2 -2)] = sqrt [ (152 + 332) /2] = 25.632
standard error of difference se0 = sp * sqrt (1/n1 + 1/n2 ) = 25.632 * sqrt [1/9 + 1/9] = 12.083
Test statistic
t = (M1 - M2 )/ se0 = (255 - 264)/ 12.083 = -0.7449
a-2 Here the critical value of t for dF = 18-2 = 16 and alpha = 0.05
tcritical = t0.10,16 = -1.337
a-3 No, since the value of test statistic is not less than the critical value.
b-1
standard error of difference = sqrt [s12 /n1 + s22 /n2] = sqrt [(332 + 152)/9] = 12.083
Here t = (M1 - M2 )/ se0 = (255 - 264)/ 12.083 = -0.7449
b-2
Here degree of freedom dF = [(s12 + s22 )2 /n2 ]/ [n2 (s14 + s24)/(n-1)n2]
dF = 21306/ 1908.25
dF = 11.165 or say 12
t11,0.1 = -1.3634
b.3 No, since the value of test statistic is not less than the critical value. Option B is correct
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