# 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

1. From the excel formulae and model, the mean of elder twins is 99.0833 and there standard deviation is 8.4689.
2. On the other hand, the mean and standard deviation for the younger twins is 102.8333 and 8.6742 respectively.
3. 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.