Respuesta :
Answer:
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce.
Step-by-step explanation:
Step:-(i)
Given first sample size n₁ = 200
The first sample proportion [tex]p_{1} = \frac{5}{200} = 0.025[/tex]
Given first sample size n₂= 500
The second sample proportion [tex]p_{2} = \frac{25}{500} = 0.05[/tex]
Step:-(ii)
Null hypothesis :H₀:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
Alternative hypothesis:-H₁
There is significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
level of significance ∝=0.10
Step:-(iii)
The test statistic
[tex]Z =\frac{p_{1} - p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]
where p = [tex]\frac{n_{1} p_{1} + n_{2}p_{2} }{n_{1}+n_{2} }= \frac{200X0.025+500X0.05 }{500+200}[/tex]
p = 0.0428
q = 1-p =1-0.0428 = 0.9572
[tex]Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571(\frac{1}{200 }+\frac{1}{500 } } }[/tex]
Z = -1.479
|z| = |-1.479|
z = 1.479
The tabulated value z= 1.645 at 0.10 or 90% level of significance.
The calculated z- value = 1.479 < 1.645 at 0.10 or 90% level of significance.
The null hypothesis is accepted at 90% level of significance.
Conclusion:-
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
