The average fruit fly will lay 404 eggs into rotting fruit. A biologist wants to see if the average will change for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. 403, 406, 390, 403, 434, 392, 416, 409, 409, 393, 432, 385, 394 What can be concluded at the the αα = 0.10 level of significance level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer < ≠ > = H1:H1: ? p μ Select an answer = > < ≠ The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer accept fail to reject reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is equal to 404. The data suggest the populaton mean is significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404. The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is not significantly different from 404 at αα = 0.10, so there is insufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404.

The average fruit fly will lay 404 eggs into rotting fruit. A biologist wants to see if the average will change for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. 403, 406, 390, 403, 434, 392, 416, 409, 409, 393, 432, 385, 394 What can be concluded at the the αα = 0.10 level of significance level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer < ≠ > = H1:H1: ? p μ Select an answer = > < ≠ The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer accept fail to reject reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is equal to 404. The data suggest the populaton mean is significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404. The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is not significantly different from 404 at αα = 0.10, so there is insufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 19PFA

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403, 406, 390, 403, 434, 392, 416, 409, 409, 393, 432, 385, 394

What can be concluded at the the αα = 0.10 level of significance level of significance?

For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
The null and alternative hypotheses would be:

H0:H0: ? μ p Select an answer < ≠ > =

H1:H1: ? p μ Select an answer = > < ≠

The test statistic ? z t = (please show your answer to 3 decimal places.)
The p-value = (Please show your answer to 4 decimal places.)
The p-value is ? > ≤ αα
Based on this, we should Select an answer accept fail to reject reject the null hypothesis.
Thus, the final conclusion is that ...
- The data suggest the population mean is not significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is equal to 404.
- The data suggest the populaton mean is significantly different from 404 at αα = 0.10, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404.
- The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is not significantly different from 404 at αα = 0.10, so there is insufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 404.

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