The average fruit fly will lay 411 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. 413, 397, 427, 426, 414, 439, 428, 421, 430, 397, 419, 440 What can be concluded at the the a - 0.01 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Họ: 2 Select an answer H: ? Select an answer c. The test statistic7 v (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 v the null hypothesis. g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) O The data suggest the population mean is not significantly different from 411 at a 0.01, 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 411. O The data suggest the populaton mean is significantly different from 411 at a - 0.01, 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 411. 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 different from 411 at a - 0.01, 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 411.

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The average fruit fly will lay 411 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.
413, 397, 427, 426, 414, 439, 428, 421, 430, 397, 419, 440
What can be concluded at the the a = 0.01 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: ? vl Select an answer v
H: ? vI 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 ? va
f. Based on this, we should Select an answer
g. Thus, the final conclusion is that ...
the null hypothesis.
O The data suggest the population mean is not significantly different from 411 at a = 0.01, 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 411.
O The data suggest the populaton mean is significantly different from 411 at a = 0.01, 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 411.
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 different from 411 at a = 0.01, 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 411.
Transcribed Image Text:The average fruit fly will lay 411 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. 413, 397, 427, 426, 414, 439, 428, 421, 430, 397, 419, 440 What can be concluded at the the a = 0.01 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: ? vl Select an answer v H: ? vI 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 ? va f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... the null hypothesis. O The data suggest the population mean is not significantly different from 411 at a = 0.01, 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 411. O The data suggest the populaton mean is significantly different from 411 at a = 0.01, 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 411. 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 different from 411 at a = 0.01, 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 411.
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