A random sample of 5 year old children is evaluated for being left- or right-handed. The sample size is n = 100 children of which 79 are right-handed, 15 are left-handed, and 6 use both hands equally.

## A random sample of 5 year old children is evaluated for being left

1. A random sample of 5 year old children is evaluated for being left- or right-handed. The sample size is n = 100 children of which 79 are right-handed, 15 are left-handed, and 6 use both hands equally. According to genetic analyses, the expression of the dominant gene for handedness is highly correlated with actual handedness. Test the following hypothesis based on the sample given above at a significance level of α = 0.05. (5 marks)

H0: p1= p2= p3= 0.33;

H1: at least two of the probabilities are not the same

need to do a chi-square anaylsis

**More details;**

The formula for a chi-square statistic is the sum of the quotient between the squares of the differences of each observed and expected frequency. In other words: http://i.imgur.com/OtGjrVR.gif Our observed frequencies are 79, 15, and 6. Our expected frequencies are 100/3 for each of the three categories.

So, our chi-square statistic is: [79-(100/3)]^2/(100/3) + [14-(100/3)]^2/(100/3) + [6-(100/3)]^2/(100/3) That equals, roughly: 61.563 + 11.213 + 22.413 = 95.189 So, our chi-square value is 95.189.

To find the p-value, we need to know our degrees of freedom. Our degrees of freedom is the number of categories minus 1, so in this case, it’s 2 (three categories minus one).

With a chi-square statistic of 95.189 and and 2 degrees of freedom, we go to our chi-square distribution chart, like the one shown here: http://passel.unl.edu/Image/Namuth-CovertDeana956176274/chi-sqaure%20distribution%20table.PNG We see that in the row with two degrees of freedom, there is only a .01 chance of a chi-square value larger than 9.21.

Our value, 95.189, is enormously larger than 9.21. That means there’s less than a .01 chance of getting our chi-square value is the proportions are all equal. (In fact, it’s less than .0001, which is the point where most calculators stop).

Thus, since our p-value is less than .01, it is also less than .05 (the significance level). Thus, we can certainly reject the null hypothesis and accept the alternative hypothesis