Question

A random sample of 220 women and 210 me coffee drinkers were questioned. The result was that 71 of the women and 58 of the men preferred decaffeinated coffee. Create a 95% Confidence interval around the difference of the means of the 2 groups. Does the data establish that the proportion of women coffee drinkers who prefer decaf coffee is the same as the corresponding proportion for men?

Answer 1

Here the proportion that women  who prefer decaf coffee is ;

x= 71

n1= 220

= 0.3227273

Similarly proportion that women  who prefer decaf coffee is ;

y= 58

n2= 210

=  0.2761905

Here to test whether  the data establish that the proportion of women coffee drinkers who prefer decaf coffee is the same as the corresponding proportion for men;

The null hypothesis will be

And the alternative hypothesis given as

Now obtaining  pooled sample proportion

p = (p1 * n1 + p2 * n2) / (n1 + n2)

=(71+58)/(220+210)

= 0.3

Now calculating

= 0.1026911

The test statistic is given as

= ( 0.3227273 - 0.2761905) / 0.1026911

= 0.4531727

Calculating the p-value for z =   0.4531727, since the test is two tail test

2P(Z > |z|)   = 0.6504

At 5 % level of significance

p-value > = 0.05

We fail to reject null hypothesis .

There is sufficient evidence that the proportion of women coffee drinkers who prefer decaf coffee is the same as the corresponding proportion for men;

Obtaining 95% confidence interval

Upper limit

=

= 0.1329874

Lower limit

= =

-0.03991384

The confidence interval is given as

(  -0.03991384,   0.1329874)

( Verifying using R-output

)