Question

26% of the general public drinks alcohol once a week. A researcher claims that adults working in “white collar” jobs have different drinking habits than the general public. Data is collected from a random sample of 120 adults with white collar jobs of which 23 stated that they drank once a week in the last month. At α = 0.01, is there enough evidence to support the researcher’s claim? Conduct the appropriate significance test.

Answer 1

Solution:

n = 120

x = 23

Hypothesis are

H₀ : p = 0.26

H_a : p 0.26

Now ,

= 23/120 = 0.1917

The test statistic z is

z =  

=  (0.1917 - 0.26)/[0.26*(1 - 0.26)/120]

= -1.71

Now ,

Here ,

Two tailed test

So ,

p value = P(Z < -z) + P(Z > +z) = P(Z < -1.71)+P(Z > +1.71) = 0.0436+0.0436 = 0.0872

Since p value is greater then  α = 0.01 ,

So , Fail to reject H₀ : p = 0.26

There is not sufficient evidence to support the claim that adults working in “white collar” jobs have different drinking habits than the general public.