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:
<u>Step:-(i)</u>
Given first sample size n₁ = 200
The first sample proportion 
Given first sample size n₂= 500
The second sample proportion 
<u>Step:-(ii)</u>
<u>Null hypothesis :H₀</u>:There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
<u>Alternative hypothesis:-H₁</u>
There is significant difference in the proportion of E. Coli in organic vs. conventionally grown produce
<u>level of significance ∝=0.10</u>
<u>Step:-(iii)</u>
The test statistic

where p = 
p = 0.0428
q = 1-p =1-0.0428 = 0.9572

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.
<u>Conclusion:</u>-
There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce