2. A researcher has a hypothesis that exposure to lead leads to reduced 1Q. They
ID: 3354727 • Letter: 2
Question
2. A researcher has a hypothesis that exposure to lead leads to reduced 1Q. They conduct lQ tests on 25 individuals from an area that is now known to have had high levels of lead in the water. Population mean is 100, X-91, variance = s2 = 37 a) Calculate the t statistic b) Using alpha 0.05, determine the critical t value, tcrit. c) what is the 95% confidence interval for the difference in the means? (hint this is a t-test CI and d) e) may use a different terit) Is there a difference in the means? What can you conclude about the effect of exposure to lead on lQ?Explanation / Answer
Solution:
Population mean= 100, population variance= 20*37=740
Mean2=91, variance= 37, n=20
t =( mean1-mean2)/ (^2/n1+^2/n2)
t = ( 100-91)/ (740/25+37/25) = 1.614
b) t(n1 + n2 -2)
=0.05
t(25+25 -2) by interpolation on table values
t( 48)= 2.0216
c) (100– 91 ) – t0.025(48) (740/25+37/25) <1 – 2 < (81 - 73 ) + t0.025(100-91) (740/25+37/25)
-2.27<1 – 2 < 10.27
d) There is NO significance in mean since 1.614<2.0216
e) We conclude that Ho is reject and H1 accepted
Ho:mean1=mean2
H1 : mean1 mean2
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