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The average fruit fly will lay 422 eggs into rotting fruit. A biologist wants to see if the average will be greater 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. 407, 439, 434, 413, 445, 424, 418, 448, 414, 428, 449, 424, 436, 425, 447 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: ? Select an answer v H1 : ? Select an answer C. The test statistic ? v = (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 ? v a 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 more than 422 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 422. The data suggest the populaton mean is significantly more than 422 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 more than 422. The data suggest that the population mean number of eggs that fruit flies with this gene