A random sample of 5 year old children is evaluated for being left- or right-han

Question

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 = 0.05.

H0: p1= p2= p3= 0.33;

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

Explanation / Answer

Here, we will use the Chi square test :

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.

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) + [15-(100/3)]^2/(100/3) + [6-(100/3)]^2/(100/3)

That equals, roughly:

61.563 + 10.083 + 22.413 = 94.059

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).
there is only a .01 chance of a chi-square value larger than 9.21. Our value, 94.059, 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.

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.

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