/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 169 Using Data 5.1 on page \(375,\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 Potatoes after 20 Days After 20 days, 250 of the 500 fruit flies eating organic potatoes are still alive, while 130 of the 500 eating conventional potatoes are still alive.

Short Answer

Expert verified
Using a z-test for proportions, you can determine whether there is a significant difference between proportions of fruit flies surviving on organic potatoes and those surviving on conventional potatoes. In this case, with \(p_{o}\) being 0.5 and \(p_{c}\) being 0.26, you will see that the proportion of fruit flies surviving on organic potatoes is significantly greater than that of those surviving on conventional potatoes at a 5% significance level.

Step by step solution

01

Formulate the hypothesis

The null hypothesis \(H_{0}\) is: \(p_{o} = p_{c}\), where \(p_{o}\) is the proportion of fruit flies alive after eating organic food and \(p_{c}\) is the proportion of fruit flies alive after eating conventional food. The alternate hypothesis, \(H_{a}\), is \(p_{o} > p_{c}\).
02

Calculate the proportions

After 20 days, 250 out of 500 fruit flies that ate organic potatoes are still alive, so the proportion, \(p_{o}\), is 250/500 = 0.5. For the fruit flies that ate conventional potatoes, 130 out of 500 are still alive, so the proportion, \(p_{c}\), is 130/500 = 0.26.
03

Conduct the hypothesis test

We want to verify whether the proportion of fruit flies surviving on organic potatoes is significantly greater than that of those surviving on conventional potatoes. To conduct this test, we will use a z-test for proportions because we have large sample sizes (n > 30). This will test if there is a significant difference between the proportions. If \(p_{o}\) is significantly greater than \(p_{c}\), we can reject the null hypothesis at the 5% significance level.

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Key Concepts

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

Proportions
Proportions play a crucial role in hypothesis testing, especially when comparing two groups like the fruit flies eating organic versus conventional potatoes. In this context, a proportion is essentially a type of percentage that shows the parts of a whole within a certain group. For example, in the given problem, we calculate proportions to find out how many fruit flies survived after being fed different types of potatoes.

To calculate the proportion in each group, you divide the number of surviving fruit flies by the total number of fruit flies in that group. In our case:
  • Organic group proportion, \(p_o\): \(\frac{250}{500} = 0.5\) or 50%. This means half of the fruit flies eating organic potatoes survived.
  • Conventional group proportion, \(p_c\): \(\frac{130}{500} = 0.26\) or 26%. So, only 26% of fruit flies eating conventional potatoes survived.

By doing this, we can visualize and quantify survival rates, forming the basis for any further statistical test, such as a Z-test, which helps us compare these proportions.
Z-Test
The Z-test for proportions is a statistical tool used when we want to know if there's a significant difference between two proportions. It is especially handy when the sample sizes are large, typically more than 30. In our exercise, we perform a Z-test to see if the survival rate for fruit flies eating organic potatoes is significantly higher than those eating conventional potatoes.

Here's how it essentially works:
  • We compare the difference between two group proportions against what we would expect if there was really no comparison to be made, meaning they were almost equal.
  • This involves calculating the standard error, which measures the variation we might expect in the sample proportions.
The Z-test gives us a Z-score which is used to determine how far and in which direction our sample statistic deviates from the null hypothesis. A larger Z-score signifies a greater deviation and potentially, a greater reason to believe in the alternate hypothesis.

For our exercise, at a 5% significance level, we would look at the Z-score to decide if \(p_o\) is greater than \(p_c\). If the Z-score exceeds a critical value, we reject the null hypothesis \((H_0)\) that says the proportions are the same.
Statistical Significance
Statistical significance is a term that tells us whether the results of our test (the Z-test, in this case) are likely due to something other than random chance. If our hypothesis testing in the fruit fly study reports statistical significance, it means the observed difference in proportions is truly significant in the context of our study.

At the heart of statistical significance is the "significance level," often denoted by alpha (\({\alpha}\)). It is the threshold at which we decide to reject the null hypothesis. A common choice is 5% or 0.05, which was also used in our fly survival exercise.

Whenever the computed p-value from a Z-test is less than this significance level, the test result is termed statistically significant. This suggests there is less than a 5% chance that the observed difference in survival proportions happened by random chance alone. Thus, proving with confidence that the organic potato diet impacts fruit fly survival rates more than the conventional diet.

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