3. Two years\' exams are sample. In year 1, a sample of ni = 18 exams had a mean

Question

3. Two years' exams are sample. In year 1, a sample of ni = 18 exams had a mean of Z1-70.3 and a standard deviation of 8.-11.8. In year 2, a sarnple of n2 = 22 exams had a mean of = 721 and a standard deviation of82-12.2. At = 0.07, was the mean of all year-1 exams actually different from the mean of all year-2 exams? (Assume that -of.) 4. In a sample of 19 restaurants, the number of smoking patrons had a variance of 14.8 At = 0.10, can we conclude that the variance of the number of smoking patrons is can we conclude that the variance of the number of smoking patrons is greater than 10.5?

Explanation / Answer

Question 3

H0 : 1 = 2

Ha ; 1 2

x1 = 70.3 ; s1 = 11.8 ; n1 = 18

x2 = 72.1 ; s1 = 12.2 ; n1 = 22

Pooled standard deviation sp = sqrt [ {(n1 -1)s12 + (n2 -1)s22 }/ (n1 + n2 -2)]

sp = sqrt [(17 * 11.82 + 21 * 12.22 )/ (17 + 21)] = 12.0227

Here standard error of difference se0 = sp * sqrt (1/n1 + 1/n2) = 12.0227 * sqrt (1/18 + 1/22) = 3.821

Here test statistid

t = (x2 - x1)/se0 = (72.1 -70.3)/ 3.821 = 0.471

Here dF = 38 and alpha = 0.07

so t38,0.07 = 1.8644

so here t > tcritical so we shall not reject the null hypothesis and can say that means for both the year examss are different.

QUestion 4

n = 19

dF = 18

Variance s2 = 14.8

Here

H0 : 2 < =10.5

Ha : 2 > 10.5

Here Test statistic

X2 = (n-1) (s/02 ) = (19 -1) * (14.8/10.5) = 25.37

so here at alpha = 0.10

X2critical = X2 0.10,18  = 25.989

so here X2 <  X2critical so we cannot reject the null hypotheiss and can conclude that number of smoking patrons is not greater than 10.5 .

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