93:22:25 A manager records the repair cost for 56 randomly selected stereos. A s

Question

93:22:25 A manager records the repair cost for 56 randomly selected stereos. A sample mean of $51.97 and standard deviation of $19.81 are subsequently computed Determine the 80% confidence interval for the mean repair cost for the stereos. Assume the population is normally distributed. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Answer(How to Enter) 4 Points Keypad Prev Next © 2017 H Learning 532 PM 3 11/2/2017 0 e- here to search

Explanation / Answer

PART A.
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 55 d.f is 1.297

PART B.
TRADITIONAL METHOD
given that,
sample mean, x =51.97
standard deviation, s =19.81
sample size, n =56
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 19.81/ sqrt ( 56) )
= 2.647
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 55 d.f is 1.297
margin of error = 1.297 * 2.647
= 3.433
III.
CI = x ± margin of error
confidence interval = [ 51.97 ± 3.433 ]
= [ 48.537 , 55.403 ]
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DIRECT METHOD
given that,
sample mean, x =51.97
standard deviation, s =19.81
sample size, n =56
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 55 d.f is 1.297
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 = [ 51.97 ± t a/2 ( 19.81/ Sqrt ( 56) ]
= [ 51.97-(1.297 * 2.647) , 51.97+(1.297 * 2.647) ]
= [ 48.537 , 55.403 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 80% sure that the interval [ 48.537 , 55.403 ] 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

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