you believe that type of vegetation that cicadas eat affects their hind tibia le
ID: 3150111 • Letter: Y
Question
you believe that type of vegetation that cicadas eat affects their hind tibia length. Namely, you believe that mature oak leaves will produce longer hind tibia lenghts compared to mulberry leaves as a food source. Therefore you randomly collect 20 young male cicades and randomly put them into two feeding of groups of 10 per group. One group of 10 is fed oak leaves while the other group of 10 is fed mulberry leaves. At maturity all hind tibia lenghts are measured with the following results
Food source
Average hind tibia length
n
S(standard deviation)
Oak leaves
79.1
10
3.13
Mulberry leaves
75.3
10
1.29
a)calculate the following confidence intervals for the mean hind tibia length for the cicades that ate oak leaves
i)90% ii)95 iii)99%
b) calculate the 95% confidece interval for the difference in average hind tibia length for the two typed of cicades(i.e those eating oak leaves and those eating mulberry leaves?
c) interpret your confidence interval in problem B above
d)construct the 95% confidence interval for the variance in hind tibia length for the cicadas eating oak leaves
e) construct the 90% confidence interval for the standard deviation for in hind tibia length for the cicades eating mulberry leaves.
Food source
Average hind tibia length
n
S(standard deviation)
Oak leaves
79.1
10
3.13
Mulberry leaves
75.3
10
1.29
Explanation / Answer
A)
i)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 79.1
t(alpha/2) = critical t for the confidence interval = 1.833112933
s = sample standard deviation = 3.13
n = sample size = 10
df = n - 1 = 9
Thus,
Margin of Error E = 1.81440218
Lower bound = 77.28559782
Upper bound = 80.91440218
Thus, the confidence interval is
( 77.28559782 , 80.91440218 ) [ANSWER]
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ii)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 79.1
t(alpha/2) = critical t for the confidence interval = 2.262157163
s = sample standard deviation = 3.13
n = sample size = 10
df = n - 1 = 9
Thus,
Margin of Error E = 2.239067116
Lower bound = 76.86093288
Upper bound = 81.33906712
Thus, the confidence interval is
( 76.86093288 , 81.33906712 ) [ANSWER]
***********************
iii)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 79.1
t(alpha/2) = critical t for the confidence interval = 3.249835542
s = sample standard deviation = 3.13
n = sample size = 10
df = n - 1 = 9
Thus,
Margin of Error E = 3.21666417
Lower bound = 75.88333583
Upper bound = 82.31666417
Thus, the confidence interval is
( 75.88333583 , 82.31666417 ) [ANSWER]
***********************
b)
Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
Calculating the means of each group,
X1 = 79.1
X2 = 75.3
Calculating the standard deviations of each group,
s1 = 3.13
s2 = 1.29
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 10
n2 = sample size of group 2 = 10
Thus, df = n1 + n2 - 2 = 18
Also, sD = 1.070560601
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.10092204
lower bound = [X1 - X2] - t(alpha/2) * sD = 1.550835638
upper bound = [X1 - X2] + t(alpha/2) * sD = 6.049164362
Thus, the confidence interval is
( 1.550835638 , 6.049164362 ) [ANSWER]
************************
c)
As 0 is not inside the interval, there is significant evidence that the true mean hind tibia length is larger for those eating oak leaves at 0.025 level.
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