MORE BENEFITS OF EATING ORGANIC
Using data from a study, 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. This exercise asks you to conduct a hypothesis test using
additional data from this study. In this case, we are testing

where and 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. 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 flies eating organic raisins are still alive, while 170 of the 500 eating conventional raisins are still
alive.
1Proportions approximated from information given in the paper.
Part 1
Give the test statistic and the p-value.
Round your answers to three decimal places.
Test statistic = i

Ask by Ingram Sandoval. in the United States

Mar 28,2025

Question

MORE BENEFITS OF EATING ORGANIC
Using data from a study, 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. This exercise asks you to conduct a hypothesis test using
additional data from this study. In this case, we are testing

where and 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. 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 flies eating organic raisins are still alive, while 170 of the 500 eating conventional raisins are still
alive.
1Proportions approximated from information given in the paper.
Part 1
Give the test statistic and the p-value.
Round your answers to three decimal places.
Test statistic = i

Ask by Ingram Sandoval. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

$$
\hat{p}_o = \frac{275}{500} = 0.55 \quad \text{and} \quad \hat{p}_c = \frac{170}{500} = 0.34.
$$

Since we assume the null hypothesis $$H_0: p_o = p_c$$, we compute the pooled proportion:

$$
\hat{p} = \frac{275+170}{500+500} = \frac{445}{1000} = 0.445.
$$

The standard error for the difference in proportions is

$$
SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_o}+\frac{1}{n_c}\right)} = \sqrt{0.445 \times 0.555 \times \left(\frac{1}{500}+\frac{1}{500}\right)}.
$$

Since

$$
\frac{1}{500}+\frac{1}{500} = \frac{2}{500} = 0.004,
$$
and

$$
0.445 \times 0.555 \approx 0.247975,
$$
we have

$$
SE = \sqrt{0.247975 \times 0.004} \approx \sqrt{0.0009919} \approx 0.0315.
$$

The test statistic is then

$$
z = \frac{\hat{p}_o - \hat{p}_c}{SE} = \frac{0.55 - 0.34}{0.0315} = \frac{0.21}{0.0315} \approx 6.667.
$$

For a one-tailed test at a $$5\%$$ significance level, the $$p$$-value is

$$
p = P(Z > 6.667).
$$

Since $$z = 6.667$$ is far in the tail of the standard normal distribution, the $$p$$-value is extremely small (essentially 0 when rounded to three decimal places).

Thus,

- Test statistic: $$ z \approx 6.667 $$
- $$ p$$-value: $$ \approx 0.000 $$
Test statistic: $$ z \approx 6.667 $$ $$ p$$-value: $$ \approx 0.000 $$

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