Question

1. Binge drinking by gender. Wechsler and coworkers (1994) found that 19.4% of students at 4-year U.S. colleges engaged in frequent binge drinking. Table 1reports the data from this study by gender.

TABLE 1. Frequency of binge drinking by gender. Binge drinking is defined as having five or more drinks in a row three or more times during the past 2-week period.

Binge+  Bing –

1630	5550
1684	8232

        Men

   Women

1. Calculate the prevalence of binge drinking for males and for females separately.
2. Fill in this blank: In absolute terms, males had a ____% greater prevalence of binge drinking than females.
3. Fill in this blank: In relative terms, males had a ____% greater prevalence of binge drinking than females.
4. Calculate a 95% confidence interval for the prevalence ratio of binge drinking for males.

Answer 1

Here we have given the genderwise 2X2 cotingency table of the induals taking binge and note taking binge as

	Binge +	Binge -	Total
Male	1630	5550	7180
Female	1684	8232	9916
Total	3314	13782	17096

Here first we have to find prevalence of binge drinking for males and for females.

The prevalece of male is denoter by ( )

  No. of male taking the bringe / Total no.of male.

The prevalece of male is 0.227 or 22.70 %

Similarly for the female

The prevalece of female is denoter by ( )

  No. of female taking the bringe / Total no.of female.

The prevalece of male is 0.170 or 17.00 %

Here we have to find the percentage increase in absolute term as

absolute measure = = 22.70 - 17.00 = 5.7 %  

Hence

In absolute terms, males had a 5.7 % greater prevalence of binge drinking than females.

Now similarly,

Here we have to find the percentage increase in relative term as

Relative measure =

that is 74.80 %    

Hence

In relative terms, males had a 74.80 % greater prevalence of binge drinking than females.

Now we have to find the 95% confidence inrterval for the prevalence of male.

The formula for the confidence interval is given by

Where   

Here is calculated from the normal table corresponding to the confidence level of 0.95 is given by

First we find the value of E .

here n = total no.of male = 7180 hence E is

therefore E = 0.009689

Hence the 95% confidence interval is

Hence the 95% confidence interval is ( 0.2173 , 0.2367 ) .