Question

An organization published an article stating that in any one-year period, approximately 10.3 percent of adults in a country suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, eight of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult population in the country.

Answer 1

Answer:

We accept H₀ we don´t have enough evidence to support that the proportion of certain town differs from that of National Information

Step-by-step explanation:

National Information:

p = 10,3 %      p = 0,103

Sample

Sample size    n₁  = 100

x₁  = 8

p₁  = x₁/100         p₁  = 8 / 100     p₁ = 0,08 %  

q₁  =  1 - p₁       q₁  = 1  - 0,08      q₁  = 0,92

Confidence Interval  CI = 95 %   significance level  α  = 5 %   α = 0,05

z(c) for α = 0,05  from z-table is  : 1,64

Test Hypothesis:

Null Hypothesis                                       H₀       p₁  = p

Alternative Hypothesis                           Hₐ       p₁ <  p

The alternative hypothesis suggest a one tail test to the left

z(s)  =  (  p₁  - p ) √p₁*q₁ / n₁

z(s)  = ( 0,08 - 0,103 )/√ 0,08*0,92/100

z(s)  =  - 0,023 / 0,027

z(s)  = - 0,85

Comparing  z(s)   and  z(c)

z(s) > z(c)     - 0,85  > -1,64

z(s) is in the acceptance region, we accept H₀