Question

A comparative study of organic and conventionally grown produce was checked for the presence of E. coli. Results are summarized below. The Prevalence of E. Coli in Organic and Conventional Produce Sample Size E. Coli Prevalence Organic 200 5 Conventional 500 25 Is there a significant difference in the proportion of E. Coli in organic vs. conventionally grown produce? Test at α = 0.10. Be sure to report your hypotheses, show all work, and explain the meaning of your answer.

Answer 1

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.

Explanation:

Step:-(i)

Given first sample size n₁ = 200

The first sample proportion
`p_(1) = (5)/(200) = 0.025`

Given first sample size n₂= 500

The second sample proportion
`p_(2) = (25)/(500) = 0.05`

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

`Z =\frac{p_(1) - p_(2) }{\sqrt{pq((1)/(n_(1) )+(1)/(n_(2) ) } }`

where p =
`(n_(1) p_(1) + n_(2)p_(2) )/(n_(1)+n_(2) )= (200X0.025+500X0.05 )/(500+200)`

p = 0.0428

q = 1-p =1-0.0428 = 0.9572

`Z =\frac{0.025- 0.05}{\sqrt{0.0428X0.9571((1)/(200 )+(1)/(500 ) } }`

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