Suppose IQ scores were obtained from randomly selected siblings. For 20 such pai

Question

Suppose IQ scores were obtained from randomly selected siblings. For 20 such pairs of people, the linear correlation coefficient is 0.927 and the equation of the regression line is y = 5.49 + 0.96x, where x represents the IQ score of the older child. Also, the 20 x values have a mean of 99.24 and the 20 y values have a mean of 101.1. What is the best predicted IQ of the younger child, given that the older child has an IQ of 107? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted IQ of the younger child is

Explanation / Answer

Answer:

First we need to understand why these values of pearson correlation coefficient, mean of X (Older child) values, mean Y (younger child) values are given. Actually,we can test or validate the acceptance of the given regression line at 5% significance level using these given values statistically. If we see that we can well accept the given regression line at 5% significance level, then we can directly use this regression line for predicting the requirement.

Here, correlation, r = 0.927 and regression coefficient, b= 0.96.

We know that b=r*Sy/Sx. Hence using value of r and b, we get Sy/Sx = 1.04 (upto 2 decimal place).

We can validate the value of Sy/Sx by its statistical distribution.Since, both the samples have equal number of observations, i.e. 20, we can say Sy/Sx will have an F-distribution with (19,19) df (since number of observations are same for both the sample, ratio of corresponding dfs will lead to 1 in Sy/Sx). From the table of F-distribution we get crtical values as 2.168 at 5% significance level. So observed value of Sy/Sx (1.04) is smaller than critical value at 5% significance. Hence we claim that Sy/Sx and hence regression coefficient is significant or acceptable at 5% significance level. Similarly if we check for intercept (,a=5.49 here), we can see from the given values of mean of both samples and regression coefficient, intercept coming out to be 5.83 (to two dp) and it is also significant at 5% level of significance (we can validate it by Z-test). So. we see that given regression line pass at 5% significance level.

Now we can put the value of X=107 in the regression equation to get the predicted IQ for Y.

Hence, the best predicetd IQ of the younger child is = 5.49+(0.96*107) = 108.21 (upto two dp)

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