Question

A local fertility specialist noticed that some of her patients were heavy coffee drinkers. She wanted to know if there was any association between coffee drinking and conception in the first year of trying. She asked a random sample 40 patients (some of whom have successfully become pregnant) about coffee drinking. The table below summarizes the data (table entries represent raw frequencies of women in terms of pregnant or not and coffee drinker or not). Using alpha =.01, use the 2-way Chi-square to determine whether there is any relation between successful pregnancy and coffee drinking.

		Coffee
		Yes (drinker)	No (no-drinker)
Pregnant	Yes (pregnant)	6	15
	Not (not pregnant)	9	10

1. Write down the null and alternative hypothesis. (2pt)

2. Specify the Degrees of Freedom (2pt).

3. Determine the critical value (need to use chi-square Table) (2pt)

4. Conduct necessary calculation and state the conclusion based on your calculation (8pt)

Answer 1

Solution:

The given data is as follows:

		Coffee
		Yes(drinker)	No(drinker)	Total
Pregnant	Yes(Pregnant)	6(a)	15(b)	21(a+b)
	Not(Pregnant)	9(c)	10(d)	19(c+d)
	Total	15(a+c)	25(b+d)	40(N)

1)Null and alternative hypothesis are as follows:

Null hypothesis is that two attributes "pregnancy and coffee drinking" are independent.

Alternative hypothesis is that they are not independent.

2)Formula to find degree of freedom is df=(rows-1)*(columns-1)=(2-1)*(2-1)=1

3)Determination of critical value

For $\alpha$ =0.01 and df=1, the critical value from Chi-square distribution table is 6.635.

Critical values of the Chi-square distribution with d degrees of freedom Probability of exceeding the critical value d 0.05 0.01 0.001 d 0.05 0.01 0.001 1 3.841 6.635 10.828 11 19.675 24.725 31.264 2 5.991 9.210 13.816 12 21.026 26.217 32.910 3 7.815 11.345 16.266 13 22.362 27.688 34.528

4)Calculation of test statistics.

Formula to find chi-square test, X²=N(ad-bc)²/ R₁R₂C₁C₂ where R₁=a+b,R₂=c+d,C₁=a+c,C₂=b+d and N=a+b+c+d

		Coffee
		Yes(drinker)	No(drinker)	Total
Pregnant	Yes(Pregnant)	6(a)	15(b)	21(a+b)
	Not(Pregnant)	9(c)	10(d)	19(c+d)
	Total	15(a+c)	25(b+d)	40(N)

From the above table let us calculate X² value.

X²=40[(6*10)-(15*9)]² /(21*19*15*25)

=225000/149625

=1.5

Since X² observed value=1.5 is less than the X² critical value=6.635,we fail to reject the null hypothesis and conclude that the two attributes "pregnancy and coffee drinking" are independent.