Pregunta
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MORE DCNEFITS OF EATING ORGANIC
Using data from a study, we find a significant difference in the proportion of frult flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. This exercise asks you to conduct a hypothesis test using addittonat ditis from this study. in this case, we are testing
where and represent the proportion of fruit flies allive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. We have . Show all remaining details in the test, using a significance level.
Effect of Organic Raisins After 20 Days
After 20 days, 275 of the 500 fruit fles eating organic raisins are still alive, while 170 of the 500 eating comventional ralsins are still alive.
Proportions approximated from information given in the paper.
Part 1
(7) Your answer is partlaily correct.
Give the test statistic and the p-value.
Round your answers to three decimal places.
Test statistic

Ask by Ford Christensen. in the United States
Mar 28,2025

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Test statistic and p-value .

Solución

The pooled proportion is calculated as
Then, the test statistic is given by
where
Substituting the values, we have
First, calculate the product in the denominator:
So,
Thus, the test statistic is
(Rounding to three decimal places, one might obtain or a value very close depending on intermediate rounding; here we use .)
The p-value for a right-tailed test is
Given that a of approximately is extremely high, the p-value is effectively
Thus, the answers are:
Test statistic and p-value .

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Beyond the Answer

To calculate the test statistic for the hypothesis test comparing the proportions of fruit flies alive after 20 days, we can use the Z-test for proportions.
  1. Calculate the sample proportions:
  2. Calculate the pooled proportion:
  3. Calculate the test statistic (Z):
  4. Calculate the p-value: Since this is a one-tailed test (we’re testing if ), we can use the standard normal distribution to find the p-value for :
The p-value is essentially zero for such a high Z-score.
So, rounding the test statistic and p-value to three decimal places, we have:
Test statistic
p-value

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