A publisher reports that 55% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 280 found that 51% of the readers owned a laptop. Make the decision to reject or fail to reject the null hypothesis at the 0.05 level.

We fail to reject the null hypothesis, we accept H₀

Step-by-step explanation:

Population proportion        p₀   =  55 %    p₀   =  0,55

Sample  size         n  =  280

Sample mean     p =  51 %            p =  0,51       q = 0,49

n*p   and  n*q   are both bigger than 10 therefore we can approximate the binomial distribution to a normal distribution

Test  Hypothesis

Null Hypothesis                             H₀                   p   =  p₀

Alternative Hypothesis                Hₐ                    p   ≠  p₀

Significance level is  α  = 0,05    but we have a two-tail test then

α/2  =  0,025

z(c)  =  - 1,96

Computing  z(s)

z(s)  =  ( p - p₀ ) / √(p*q)/n

z(s)  =  ( 0,51 - 0,55 ) / √(0,51*0,49)/280

z(s)  =  - 0,04 * 16,73 / 0,5

z(s)  =  -  1,34

Comparing

z(s)   and z(c)

- 1,34    <    - 1,96

| z(s) | < | z(c)|

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

chicken nugget stir fry

step-by-step explanation: