The average fruit fly will lay 408 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. 383, 412, 400, 401, 400, 416, 389, 401, 413, 389, 400, 388, 376, 389, 412, 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 t-test for a population mean z-test for a population proportion  The null and alternative hypotheses would be:       H0:H0:  ? p μ  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 reject fail to reject  the null hypothesis. Thus, the final conclusion is that ... 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 408 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 408. The data suggest the population mean is not significantly different from 408 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 408. The data suggest the populaton mean is significantly different from 408 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 408.

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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The average fruit fly will lay 408 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.

383, 412, 400, 401, 400, 416, 389, 401, 413, 389, 400, 388, 376, 389, 412, 394

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

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

 H0:H0:  ? p μ  Select an answer < ≠ = >       

 H1:H1:  ? p μ  Select an answer > ≠ < =    

  1. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
  2. The p-value =  (Please show your answer to 4 decimal places.)
  3. The p-value is ? > ≤  αα
  4. Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.
  5. Thus, the final conclusion is that ...
    • 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 408 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 408.
    • The data suggest the population mean is not significantly different from 408 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 408.
    • The data suggest the populaton mean is significantly different from 408 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 408.
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