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The average fruit fly will lay 389 eggs into rotting fruit. A biologist wants to see if the average will be fewer 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. 358, 388, 362, 382, 371, 385, 384, 393, 403, 389, 405 What can be concluded at the the a = 0.10 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho ? ◇ Select an answer H₁: ? ◇ Select an answer c. The test statistic ?> your answer to 3 decimal places.) d. The p-value = to 4 decimal places.) e. The p-value is ? ✨ a (please show (Please show your answer f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 389 at a = 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 389. 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 less than 389 at a = 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 less than 389. The data suggest the populaton mean is significantly less than 389 at a = 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 less than 389.

The average fruit fly will lay 389 eggs into rotting fruit. A biologist wants to see if the average will be fewer 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. 358, 388, 362, 382, 371, 385, 384, 393, 403, 389, 405 What can be concluded at the the a = 0.10 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho ? ◇ Select an answer H₁: ? ◇ Select an answer c. The test statistic ?> your answer to 3 decimal places.) d. The p-value = to 4 decimal places.) e. The p-value is ? ✨ a (please show (Please show your answer f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 389 at a = 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 389. 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 less than 389 at a = 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 less than 389. The data suggest the populaton mean is significantly less than 389 at a = 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 less than 389.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The average fruit fly will lay 389 eggs into rotting fruit. A biologist wants to see if the average will be fewer 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. 358, 388, 362, 382, 371, 385, 384, 393, 403, 389, 405 What can be concluded at the the a = 0.10 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho ? ◇ Select an answer H₁: ? ◇ Select an answer c. The test statistic ?> your answer to 3 decimal places.) d. The p-value = to 4 decimal places.) e. The p-value is ? ✨ a (please show (Please show your answer f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 389 at a = 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 389. 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 less than 389 at a = 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 less than 389. The data suggest the populaton mean is significantly less than 389 at a = 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 less than 389.
Transcribed Image Text:The average fruit fly will lay 389 eggs into rotting fruit. A biologist wants to see if the average will be fewer 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. 358, 388, 362, 382, 371, 385, 384, 393, 403, 389, 405 What can be concluded at the the a = 0.10 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho ? ◇ Select an answer H₁: ? ◇ Select an answer c. The test statistic ?> your answer to 3 decimal places.) d. The p-value = to 4 decimal places.) e. The p-value is ? ✨ a (please show (Please show your answer f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 389 at a = 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 389. 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 less than 389 at a = 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 less than 389. The data suggest the populaton mean is significantly less than 389 at a = 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 less than 389.
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