A researcher has a hypothesis that exposure to lead leads to reduced IQ. They co
ID: 3203934 • Letter: A
Question
A researcher has a hypothesis that exposure to lead leads to reduced IQ. They conduct IQ 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 = s^2 = 37, n 20 Calculate the t statistic Using alpha 0.05, determine the critical t value, t_crit. What is the 95 percentage confidence interval for the difference in the means? Is there a difference in the means? What can you conclude about the effect of exposure to lead on lQ?Explanation / Answer
Answer
Given,
=0.05
Population mean(µ)=100
Sample mean(x)=91
Variance(s2)=37
Standard deviation(s)=6.082
n=25
A.a) t-statistic(t)=[x- µ]/s/sqrt(n)
t=[91-100]/6.082/5
t=-7.398
A.b) =0.05
n=25
df=25-1=24
t(crit)=t(0.025,24)=2.064
A.c) confidence interval is given by X±t*s/sqrt(n)
CI=91±2.064*6.082/5
CI=91±2.51
88.49 CI 93.51
A.d) Null hypothesis:H0 µ1-µ2=0
Alternate hypothesis:H1 µ1-µ20
t-statistic=-7.398
t(crit)=2.064
as t-statistic is beyond the critical region, we reject the null hypothesis
hence, we can say that there is difference in the means
A.e) from the hypothesis we can conclude that exposure of lead leads to reduced IQ
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