Two independent random samples from two normal populations produced the followin

Question

Two independent random samples from two normal populations produced the following summery. Our aim is to investigate if the mean of population 1 is less than the mean of the second populations. a State the null and alternative hypotheses. Suppose alpha = 0.01 Find the rejection region (use critical value). b. Find the value of the test statistic A random sample of size n = 460 is drawn from a population. This sample produced X = 345 successes. Given H_o: pi = 0.6 versus H_a: pi > 0.6, find the value of the test statistic and the p-value. Two neighborhoods (A and B) in a big city are being considered for after school programs. One hundred parents were sampled form neighborhood A and found that eighty worked full-time. Of the one hundred fifty from neighborhood one B, one hundred worked full-time. Our aim is to investigate if there is a difference between the proportion of working parents in neighborhood A and B. State the null alternative hypotheses. If alpha = .02, find the rejection region.

Explanation / Answer

Q . 1

Null HYpothesis : H0 : 1 = 2

ALternative Hypothesis : HA : 1 < 2

so it is right tailed t - test

Sample 1 Mean x1 = 21.3 ; standard deviation s12 = 12.5 and n1 = 11

Sample 2 Mean x2 = 26.5 ; standard deviation s12 = 9.5 and n1 = 19

Test Statistic : Here variance of both populations are equal

so, sp2 = [(n1 -1) s1 2 +  (n2 -1) s2 2] / (n1 + n2 -2) = 10.5714

t- stat => t = (x2 - x1 )/ sqrt (sp 2 (1/n1 + 1/n2) ) = ( 26.5 - 21.3)/sqrt (10.5714 * (1/11 + 1/19) = 5.2 / 1.2318 = 4.22

for dF = 28, tcritical = 2.467

so as t > tcritical , we can reject the null hypothesis. and can say that mean 1 islesser than mean 2.

Q.4 n = 460 ; X = 345 success

so p = 345/460 = 0.75 ; p0 = 0.6

null Hypothesis : H0 : p = 0.6

alternative hypothesis : H1 : p > 0.6

t - statistic

z = (p - 0.6)/ sqrt(0.6 * 0.4 * 460) = (0.75 - 0.6)/ sqrt (0.6 * 0.4/ 460) = 0.15/ 0.02284 = 6.567

P - value = 0.0001

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