Question

The average fruit fly will lay 391 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. 394, 390, 409, 405, 385, 359, 359, 391, 380, 407, 378 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: ? ? Hj: ? ? c. The test statistic ? = (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) 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 ... The data suggest the populaton mean is significantly less than 391 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 391. The data suggest the population mean is not significantly less than 391 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 391. 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 391 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 391.

Answer 1

meam (Ž= x = 4257 = 1387 Sample Standard deviation (s) = [(x-x)? n- = 394-387) * -- +/378–38772 11-1 2984 10 = 12984 = (17.2743 h=11, = 387, S = 17.9743, 2= dilo Hoid=391 Hi: 4 < 391 (one failed text) is unkown So we use population Standard deviation the t-distribution. Test statistic, pell. slñ . t= -0768 387-391 =-0.767991 17:2743) JT

-11-) 10 P-value = Po(t <-0.768) difanai A = [0-2301 ggel fusing P-value] ! I je l Glaulator J The result is not significant at P <d=0:10 Since, P value is greader than the d=0010 So, we do not reject the Ho Conclusion. The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in notting fruit is not significantly less than 391 at a=0.10, so there is sufficient evidence to conclude that the population mean number y eggs that fruit flies with this gene modified will lay in setting fruit is less than 39.