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 of 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 hypotheses.

2. Specify the Degrees of Freedom.

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

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

Answer 1

Ho: given two variable are independent
H1: Given two variables are not independent

	Observed Frequencies
	0
0	yes	no	F	P			Total
yes	6	9					15
no	15	10					25

	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	yes	no	F	P			Total
yes	21*15/40=7.875	19*15/40=7.125					15
no	21*25/40=13.125	19*25/40=11.875					25

	(fo-fe)^2/fe
yes	0.446	0.493
no	0.268	0.296

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   1.504                      
                          
Level of Significance =   0.01                      
Number of Rows =   2                      
Number of Columns =   2                      
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 ) =   1                      
                          
  
Critical Value =   6.635   [ Excel function: =CHISQ.INV.RT(α,DF) ]                  
Decision: test statistic,X² < critical value , So, Do not reject the null hypothesis     

Means no association.

THANKS

revert back for doubt

please upvote