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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.

Step by step solution

01

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 \(p_c = \frac{320}{500} = 0.64\).

02

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.

03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Comparison

Proportion Comparison is a method used in statistics to assess differences between two groups based on their categorical data. In this scenario, it helps us determine if there's a significant difference in fruit fly survival rates when fed different types of bananas over a period.
The core idea behind comparing proportions is to ascertain whether the observed proportions in two samples differ due to random chance or if there is a significant difference present. When we discuss proportions, we're referring to a fraction or percentage of a specific outcome within a group. Here, a proportion tells us the part of the fruit fly group that survived.

• For the organic banana group, the survival proportion is found by dividing the number of surviving flies by the total number (e.g., 345 out of 500 flies).
• The conventional group’s proportion is calculated similarly (e.g., 320 out of 500).

These proportions are then compared to understand the difference in effect due to the type of food, paving the way for further statistical testing.

Z-Test

A Z-Test is a statistical method used when testing hypotheses about proportions. It's especially useful when dealing with large sample sizes, like in our fruit fly experiment.A Z-Test helps determine if the difference in observed proportions between two groups is statistically significant.The way we conduct a Z-Test begins with calculating the Z-score, which tells us how many standard deviations an element is from the mean. Here's how it works in our example with fruit flies:

• First, we determine the known sample proportions of both groups as explained previously.
• Next, we calculate the pooled sample proportion, \[\hat{p} = \frac{n_{1}p_{o} + n_{2}p_{c}}{n_{1} + n_{2}}\]
• The Z-score is then computed using the formula:\[Z = \frac{(p_{o} - p_{c}) - 0}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_{1}} + \frac{\hat{p}(1-\hat{p})}{n_{2}}}}\]The calculated Z-score lets us assess how likely the difference between group proportions is occurring by chance.

The higher the Z-score, in this case derived as 2.2, the more likely the observed difference is significant.

Significance Level

The Significance Level is an important concept in hypothesis testing. It’s the threshold we set for deciding when to reject the null hypothesis. Often denoted as \(\alpha\), the significance level represents the probability of rejecting the null hypothesis when it is actually true.In our fruit fly test, a significance level of \(5\%\), or \( \alpha = 0.05 \), was used. This means that there's a 5% risk of concluding there is an effect when there is none.
To make decisions based on the Z-Test, the calculated Z-score is compared to a Z critical value corresponding to the chosen significance level:

• If our calculated Z-score is higher than this critical value (1.645 for a 5% level), we conclude that the difference in proportions is statistically significant – like we did in our case, leading to rejecting the null hypothesis.

Understanding significance levels ensures that the conclusions drawn from statistical tests are reliable and reflect true differences, not just random variability.