Question

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Bananas after 15 Days After 15 days, 345 of the 500 fruit flies eating organic bananas are still alive, while 320 of the 500 eating conventional bananas are still alive.

Short answer

The null hypothesis is rejected. There is a significant difference in the survival rates of flies after 15 days between those consuming organic bananas and those eating conventional bananas.

Step-by-step solution

Calculate the sample proportions

First we need to calculate the sample proportions. The proportion of fruit flies that survived in the organic group is given by \(p_o = \frac{345}{500} = 0.69\). Similarly, the proportion of fruit flies that survived in the conventional group is \(p_c = \frac{320}{500} = 0.64\).

Calculate the test statistic

The test statistic in this case is computed as \[Z = \frac{(p_o - p_c) - 0}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1} + \frac{\hat{p}(1-\hat{p})}{n_2}}}\] where \(\hat{p} = \frac{n_o p_o + n_c p_c}{n_o + n_c}\). Let’s calculate the pooled sample proportion first: \(\hat{p} = \frac{500*0.69 + 500*0.64}{500 + 500} = 0.665\). Then, plug in the values into the Z-equation: \(Z = \frac{0.69 - 0.64}{\sqrt{0.665 * 0.335 * (\frac{1}{500} + \frac{1}{500})}}\) This will give the Z value.

Make decision

Now we compare the result to the z critical value for a 5% significance level, which is approximately 1.645. If the computed z score is greater than 1.645, we reject the null hypothesis. We calculate the value to be approximately 2.2, therefore, we reject the null hypothesis and conclude there's significant evidence at the 5% level to claim more flies survive eating organic bananas after 15 days than those eating conventional bananas.