Question

Using the following data on credit scores [634, 635, 598, 511, 579, 585, 659, 412, 691, 575, 400], find the following interval estimates for a mean or median using a 95% confidence.
a. Parametric
b. Robust
c. Nonparametric - Walsh Averages

Answer 1

following data on credit scores [634, 635, 598, 511, 579, 585, 659, 412, 691, 575, 400]
so that given dat we need to find parametric test such as unpaired,paired t test.
now we are using unpaired t test.
95% of confidence interval for mean.

TRADITIONAL METHOD
given that,
sample mean, x =570.8182
standard deviation, s =94.5175
sample size, n =11
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 94.5175/ sqrt ( 11) )
= 28.498
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 10 d.f is 2.228
margin of error = 2.228 * 28.498
= 63.494
III.
CI = x ± margin of error
confidence interval = [ 570.8182 ± 63.494 ]
= [ 507.324 , 634.312 ]
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DIRECT METHOD
given that,
sample mean, x =570.8182
standard deviation, s =94.5175
sample size, n =11
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 10 d.f is 2.228
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 = [ 570.8182 ± t a/2 ( 94.5175/ Sqrt ( 11) ]
= [ 570.8182-(2.228 * 28.498) , 570.8182+(2.228 * 28.498) ]
= [ 507.324 , 634.312 ]
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interpretations:
1) we are 95% sure that the interval [ 507.324 , 634.312 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean