In: Statistics and Probability
In the North American population, the average IQ is 100. A team of scientists wants to test a new medication to see if it has either a positive or negative effect on intelligence or no effect at all. A sample of 39 participants who have taken the medication has been investigated and their IQ after treatment is noted as following:
95 90 100 110 108 105 98 105 120 115 99 100 90 118 112 110 120 120 95 100 90 110 100 105 100 90 105 103 108 112
114 100 88 95 105 103 105 107 98
1-Does the medication influence the North American population IQ? Explain and justify your answer.
The same medication is used on Scandinavian population. 20 participants have been treated and their IQs are noted.
112 114 100 88 95 105 108 105 98 105 120 115 99
100 90 112 110 120 120 90
2-Does the medication have the same effect on both samples (North American and Scandinavian) IQ? Explain and justify your answer.
Here, under normality assumption with equal variances for two independent samples, we have performed the folloeing tests:
For the first case R shows p-value=0.0105<0.05. hence we reject the null and can say that there is enough evidence to support the claim that the medication affects the North Americal population IQ, more specifically it increases its level(evident from 95% CI)
For the second case R shows p-value=0.03076<0.05. hence here also we reject the null and can say that there is enough evidence to support the claim that the medication affects the scandanavian population IQ, more specifically it increases its level(evident from 95% CI) in this case also.
For the third case we want to know whether this medication results in increment of the average IQ upto the same level for both the North American and the Scandanavian population IQ. Here p-value=0.5575>>0.05, so do not reject the null. Hence we can conclude that after this medication, both population end with same IQ level(after increment), on an average.
R CODE and OUTPUT:
>
x=c(95,90,100,110,108,105,98,105,120,115,99,100,90,118,112,110,
+ 120,120,95,100,90,110,100,105,100,90,105,103,108,112,
+ 114,100,88,95,105,103,105,107,98)
>
t.test(x,mu=100,alternative="two.sided",conf.level=0.95)
One Sample t-test
data: x
t = 2.6923, df = 38, p-value = 0.0105
alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
100.9414 106.6483
sample estimates:
mean of x
103.7949
>
y=c(112,114,100,88,95,105,108,105,98,105,120,115,99,
+ 100,90,112,110,120,120,90)
>
t.test(y,mu=100,alternative="two.sided",conf.level=0.95)
One Sample t-test
data: y
t = 2.3336, df = 19, p-value = 0.03076
alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
100.5463 110.0537
sample estimates:
mean of x
105.3
>
>
t.test(x,y,mu=0,alternative="two.sided",var.equal=TRUE,paired=FALSE,conf.level=0.95)
Two Sample t-test
data: x and y
t = -0.58997, df = 57, p-value = 0.5575
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-6.613780 3.603523
sample estimates:
mean of x mean of y
103.7949 105.3000