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Recent incidents of food contamination have caused great concern among consumers. An article reported that 31 of 80 randomly selected Brand A brand chickens tested positively for either campylobacter or salmonella (or both), the leading bacterial causes of food-borne disease, whereas 61 of 80 Brand B brand chickens tested positive.

Required:
a. Does it appear that the true proportion of non-contaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level 0.01 by obtaining a P-value.
b. If the true proportions of non-contaminated chickens for the Perdue and Tyson brands are 0.50 and 0.25, respectively, how likely is it that the null hypothesis of equal proportions will be rejected when a 0.01 significance level is used and the sample sizes are both 80?

Respuesta :

Answer:

a) z(s)  is in the rejection region for H₀  we reject H₀.

We have enough support to conclude that the two populations are different  

z(s) is in the rejection region so we reject H₀

The two population have a different proportion of non-contaminated food

Step-by-step explanation:

Conditions:

-Independent samples

-Random samples

- Size samples big enough

a) A Brand Chickens Sample:

sample size   n₁  = 80

We will go through a z- test

number of contamination cases  x₁  =  31

Then   p₁  =  31/80     p₁  =  0.3875    and  q₁  =  1 - p₁  ; q₁ = 1 - 0.3875

q₁  =  0.6125

B Brand Chickens Sample:

sample size   n₂  =  80

number of contamination cases  x₂  =  61

Then  p₂ = 61/80   p₂ = 0.7625  and  q₂ =  1  -  p₂ ;   q₂  =  1  -  0.7625

q₂  =  0.2375

Hypothesis Test:

Null Hypothesis                                  H₀     p₂ =  p₁

Alternative Hypothesis                      Hₐ     p₂ ≠ p₁

The alternative hypothesis call for a two-tail test ( ≠ ) we will test if the proportion differs )

A significance level  α =  0.01   then    α/2 = 0.005

From z -Table     α/2  =  0.005  correspond to z(c) =  2.575

To calculate z(s)

z(s)  =   ( p₂  -  p₁ ) / √( p₁*q₁/n₁) + (p₂*q₂)/n₂)

z(s)  = (  0.7625  -  0.3875 ) / √(0,3875*0.6125)/80 + (0.7625*0.2375)/80

z(s)  = ( 0.375 ) / √(0.00296) + (0.00226)

z(s)  = ( 0.375 ) /0.072

z(s)  =  5.21

Comparing    z(s)   and   z(c)    z(s) > z(c)

z(s)  is in the rejection region for H₀  we reject H₀.

We have enough support to conclude that the two populations are different

b) folowing the same procedure now talking about non-contaminated chickens

Tyson  Brand Chickens  p₁ =  0,5   then  q₁ = 0.5

Perdue Brand chickens  p₂ = 0.25 then  q₂ = 0.75

n₁ = n₂ = 80

Hypothesis Test:

Null Hypothesis                        H₀            p₁  = p₂

Alternative Hypothesis            Hₐ            p₁ ≠ p₂

Alternative hypothesis indicates a two-tail test

z(c)  =  2.575

And z(s)  =  (0.5 - 0.25)/ √(0.5*0.5)/80  + (0.25*0.75)/80

z(s) = 0.25 / √( 0.003125 + 0.002343

z(s) = 0.25 / 0.0739

z(s) = 3.38

Comparing z(s)  and z(c)

z(s) >  z(c)  

z(s) is in the rejection region so we reject H₀

The two population have different proportion of non-contaminated food

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