Question 7 0/1 pt 399 Details The average fruit fly will lay 395 eggs into rotting...

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Question 7 0/1 pt 399 Details The average fruit fly will lay 395 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. 410, 362, 390, 369, 402, 412, 363, 411, 409, 387, 408, 402, 373, 392 What can be concluded at the the ax = 0.05 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 H: ?V Select an answer c. The test statistic? d. The p-value= e. The p-value is (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.) f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... the null hypothesis. The data suggest the populaton mean is significantly different from 395 at ax = 0.05, 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 395. The data suggest the population mean is not significantly different from 395 at a = 0.05, 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 395. 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 395 at a = 0.05, 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 395. Question 7 0/1 pt 399 Details The average fruit fly will lay 395 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. 410, 362, 390, 369, 402, 412, 363, 411, 409, 387, 408, 402, 373, 392 What can be concluded at the the ax = 0.05 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 H: ?V Select an answer c. The test statistic? d. The p-value= e. The p-value is (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.) f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... the null hypothesis. The data suggest the populaton mean is significantly different from 395 at ax = 0.05, 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 395. The data suggest the population mean is not significantly different from 395 at a = 0.05, 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 395. 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 395 at a = 0.05, 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 395.