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normal. 362, 395, 367, 389, 403, 374, 378, 404, 362, 399, 382, 383, 399, 406, 386, 390 What can be concluded at the the a= 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: ? H₁: ? Select an answer Select an answer c. The test statistic? C that had this gene modified. Assume that the distribution of the popu d. The p-value e. The p-value is ? a = (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.)

normal. 362, 395, 367, 389, 403, 374, 378, 404, 362, 399, 382, 383, 399, 406, 386, 390 What can be concluded at the the a= 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: ? H₁: ? Select an answer Select an answer c. The test statistic? C that had this gene modified. Assume that the distribution of the popu d. The p-value e. The p-value is ? a = (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 10CYU
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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. 362, 395, 367, 389, 403, 374, 378, 404, 362, 399, 382, 383, 399, 406, 386, 390 What can be concluded at the the a = 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: ? H₁: ? Select an answer Select an answer c. The test statistic = (please show your answer to 3 decimal places.) d. The p-value = e. The p-value is ? a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) O The data suggest the populaton mean is significantly less than 389 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 less than 389. O 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 = 0.05, 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 less than 389. The data suggest the population mean is not significantly less than 389 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 389. Hint: Helpful Videos: Calculations [+] Setup [+] Interpretations [+] Help
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. 362, 395, 367, 389, 403, 374, 378, 404, 362, 399, 382, 383, 399, 406, 386, 390 What can be concluded at the the a = 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: ? H₁: ? Select an answer Select an answer c. The test statistic = (please show your answer to 3 decimal places.) d. The p-value = e. The p-value is ? a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) O The data suggest the populaton mean is significantly less than 389 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 less than 389. O 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 = 0.05, 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 less than 389. The data suggest the population mean is not significantly less than 389 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 389. Hint: Helpful Videos: Calculations [+] Setup [+] Interpretations [+] Help
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