Question

In a survey, 32 people were asked how much they spent on their child's last birthday gift. The results were roughly shaped as a normal curve with a mean of $39 and standard deviation of $6. Find the margin of error at a 98% confidence level.

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =39
standard deviation, s =6
sample size, n =32
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6/ sqrt ( 32) )
= 1.061
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 31 d.f is 2.453
margin of error = 2.453 * 1.061
= 2.602
III.
CI = x ± margin of error
confidence interval = [ 39 ± 2.602 ]
= [ 36.398 , 41.602 ]
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DIRECT METHOD
given that,
sample mean, x =39
standard deviation, s =6
sample size, n =32
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 31 d.f is 2.453
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 = [ 39 ± t a/2 ( 6/ Sqrt ( 32) ]
= [ 39-(2.453 * 1.061) , 39+(2.453 * 1.061) ]
= [ 36.398 , 41.602 ]
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interpretations:
1) we are 98% sure that the interval [ 36.398 , 41.602 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
Answer:
margin of error = 2.602