Effect of Heredity and Environment on IQ

Effect of Heredity and Environment on IQ

 

	x	y
	107	111
	96	97
	103	116
	90	107
	96	99
	113	111
	86	85
	99	108
	109	102
	105	105
	96	100
	89	93
Mean	99.08333	102.8333
Standard Deviation	8.468748	8.674239

 

99. From the excel formulae and model, the mean of elder twins is 99.0833 and there standard deviation is 8.4689.
100. On the other hand, the mean and standard deviation for the younger twins is 102.8333 and 8.6742 respectively.
101. Hypothesis testing

Null hypothesis: there is no difference in means (mean of x is equal to the mean of y).

Alternate hypothesis: there exists a difference in means (mean of x is equal to the mean of y)

1. A sample t test is used

 

t-Test: Two-Sample Assuming Unequal Variances
	Variable 1	Variable 2
Mean	99.08333	102.8333
Variance	71.7197	75.24242
Observations	12	12
Hypothesized Mean Difference	0
df	22
t Stat	-1.07157
P(T<=t) one-tail	0.147765
t Critical one-tail	1.717144
P(T<=t) two-tail	0.29553
t Critical two-tail	2.073873

 

N1=n2=n3

Let the n1 be sample size from population 1. S1 be the sample standard deviation of population 1. Let n2 be the sample size from population 2, s2 be the standard deviation of population 2. The common standard deviation the common standard deviation can be estimated by the pooled standard deviation.

The test statistic is:

With degree of freedom equal to df=n1+n2-2

N1=n2 = 12

Therefore t = 1.094.

With the critical t value at 0.005 significance level and 23 degree of freedom  (n1+n2-1) 1.714.

Therefore, the null hypothesis is accepted.

1. The result indicates that there is no significance difference from the t test.