Question

The following values were sampled independently from F(x).
0.86 0.48 4.43 0.87 2.51

Find the empirical cdf for these data, and add 80% confidence band limits. Summarize the information in a table with columns for the estimated cdf Fˆ(x), the range for x, and the lower and upper band limits.

Answer 1

Mean = Sum of observations/ Count of observations
Mean = (0.86 + 0.48 + 4.43 + 0.87 + 2.51 / 5) = 1.83
Variance
Step 1: Add them up
0.86 + 0.48 + 4.43 + 0.87 + 2.51 = 9.15
Step 2: Square your answer
9.15*9.15 =83.7225
…and divide by the number of items. We have 5 items , 83.7225/5 = 16.7445
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
0.86^2 + 0.48^2 + 4.43^2 + 0.87^2 + 2.51^2 = 27.6519
Step 4: Subtract the amount in Step 2 from the amount in Step 3
27.6519 - 16.7445 = 10.9074
Step 5: Subtract 1 from the number of items in your data set, 5 - 1 = 4
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
10.9074 / 4 = 2.7269
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
1.6513

a.
TRADITIONAL METHOD
given that,
sample mean, x =1.83
standard deviation, s =1.6513
sample size, n =5
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.6513/ sqrt ( 5) )
= 0.738
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 4 d.f is 1.533
margin of error = 1.533 * 0.738
= 1.132
III.
CI = x ± margin of error
confidence interval = [ 1.83 ± 1.132 ]
= [ 0.698 , 2.962 ]
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DIRECT METHOD
given that,
sample mean, x =1.83
standard deviation, s =1.6513
sample size, n =5
level of significance, α = 0.2
from standard normal table, two tailed value of |t α/2| with n-1 = 4 d.f is 1.533
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 1.83 ± t a/2 ( 1.6513/ Sqrt ( 5) ]
= [ 1.83-(1.533 * 0.738) , 1.83+(1.533 * 0.738) ]
= [ 0.698 , 2.962 ]
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interpretations:
1) we are 80% sure that the interval [ 0.698 , 2.962 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population mean
Answer:
mean = 1.83
sample size is 5
standard deviation = 1.6513
range is 4.43- 0.48 =3.95
we are 80% sure that the interval [ 0.698 , 2.962 ]