Question

The average fruit fly will lay 400 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. 410, 415, 411, 404, 369, 395, 408, 374, 398, 386, 379 What can be concluded at the the α = 0.01 level of significance level of significance? For this study, we should use Select an answer The null and alternative hypotheses would be: H 0 : ? Select an answer H 1 : ? Select an answer The test statistic ? = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? α Based on this, we should Select an answer the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton mean is significantly different from 400 at α = 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 400. 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 400 at α = 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 400. The data suggest the population mean is not significantly different from 400 at α = 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 400. Submit Question Question 11

Answer 1

Given that,
population mean(u)=400
sample mean, x =395.363
standard deviation, s =15.352
number (n)=11
null, Ho: μ=400
alternate, H1: μ!=400
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =3.169
since our test is two-tailed
reject Ho, if to < -3.169 OR if to > 3.169
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =395.363-400/(15.352/sqrt(11))
to =-1.002
| to | =1.002
critical value
the value of |t α| with n-1 = 10 d.f is 3.169
we got |to| =1.002 & | t α | =3.169
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.0018 ) = 0.3401
hence value of p0.01 < 0.3401,here we do not reject Ho
ANSWERS
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null, Ho: μ=400
alternate, H1: μ!=400
test statistic: -1.002
critical value: -3.169 , 3.169
decision: do not reject Ho
p-value: 0.3401
we do not have enough evidence to support the claim that population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 400.