Question

Recent incidents of food contamination have caused great concern among consumers. An article reported that 33 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 69 of 80 Brand B brand chickens tested positive.

(a)

Does it appear that the true proportion of non-contaminated Brand A chickens differs from that for Brand B? Carry out a test of hypotheses using a significance level 0.01. (Use p₁ for Brand A and p₂ for Brand B.)

State the relevant hypotheses.

H₀: p₁ − p₂ = 0
H_a: p₁ − p₂ < 0 H₀: p₁ − p₂ > 0
H_a: p₁ − p₂ = 0     H₀: p₁ − p₂ = 0
H_a: μ₁ − μ₂ > 0 H₀: p₁ − p₂ < 0
H_a: p₁ − p₂ = 0 H₀: p₁ − p₂ = 0
H_a: p₁ − p₂ ≠ 0

Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z = P-value =

State the conclusion in the problem context.

Reject H₀. The data suggests that the true proportion of non-contaminated chickens differs for the two companies. Fail to reject H₀. The data does not suggest that the true proportion of non-contaminated chickens differs for the two companies.     Reject H₀. The data does not suggest the true proportion of non-contaminated chickens differs for the two companies. Fail to reject H₀. The data suggests the true proportion of non-contaminated chickens differs for the two companies.

(b)

If the true proportions of non-contaminated chickens for the Brand A and Brand B 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 60? (Round your answer to four decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Recent incidents of food contamination have caused great concern among consumers. An article reported that 33 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 69 of 80 Brand B brand chickens tested positive. (a) Does it appear that the true proportion of non-contaminated Brand A chickens differs from that for Brand B? Carry out a test of hypotheses using a significance level 0.01. (Use P1 for Brand A and P2 for Brand B.) State the relevant hypotheses. O Ho: P1 - P2 = 0 Ha: P1 - P2 < 0 O Ho: P1 - P2 > 0 Ha: Pi - P2 = 0 O Ho: P1 - P2 = 0 Ha: M1 - U2 > 0 Ho: P1 – P2 < 0 Ha: P1 – P2 = 0 O Ho: P1 - P2 = 0 Ha: P1 - P20 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your p-value to four decimal places.) Z = P-value = State the conclusion in the problem context. O Reject Ho. The data suggests that the true proportion of non-contaminated chickens differs for the two companies. Fail to reject Ho. The data does not suggest that the true proportion of non-contaminated chickens differs for the two companies. Reject Ho. The data does not suggest the true proportion of non-contaminated chickens differs for the two companies. O Fail to reject Ho. The data suggests the true proportion of non-contaminated chickens differs for the two companies. (b) If the true proportions of non-contaminated chickens for the Brand A and Brand B 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 60? (Round your answer to four decimal places.)

Answer 1

1)

O Ho: P1 - P2 = 0 Ha: P1 - P270

	1st group	2nd group
x=	33	69
n =	80	80
p̂=x/n=	0.4125	0.8625
estimated prop. diff =p̂1-p̂2    =		-0.4500
pooled prop p̂ =(x1+x2)/(n1+n2)=		0.6375
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) =		0.0760
test stat z=(p̂1-p̂2)/Se =		-5.92
P value   =	0.0000	(from excel:2*normsdist(-5.92)

O Reject Ho. The data suggests that the true proportion of non-contaminated chickens differs for the two companies.

b_)

p1=		0.5
p2=		0.25
n1=		60
n2=		60
pooled proportion p=(n1p1+n2p2)/(n1+n2)=		0.375
Se(pooled)=√(p1*(1-p1)*(1/n1+1/n2) =		0.0884
Se(p1^-p2^) =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =		0.0854
for 0.01 level with two tailed test , critical zα=		2.576	(from excel:normsinv(0.005)
z score=(-/+zα*Se(pooled)-(p1-p2))/Se(p1^-p2^)=		-5.5939 and -0.2615
type II error=P(-5.5939<Z<-0.2615)=0.3969
Power =1-type II error =1-0.3969=0.6031