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alexgriva [62]
2 years ago
3

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.
Mathematics
1 answer:
MaRussiya [10]2 years ago
4 0

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     p_{1} = \frac{5}{200} = 0.025

Given first sample size n₂= 500

The second sample proportion     p_{2} = \frac{25}{500} = 0.05

<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    

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

     where         p =   \frac{n_{1} p_{1} + n_{2}p_{2}  }{n_{1}+n_{2}  }= \frac{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(\frac{1}{200 }+\frac{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.

<u>Conclusion:</u>-

There is no significant difference in the proportion of E. Coli in organic vs. conventionally grown produce

                                 

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