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.

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jtellezd jtellezd · Answer 1 · 2021-04-22T05:23:24+02:00

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