(1 point) How much money do winners go home with from the television quiz show J
ID: 3332724 • Letter: #
Question
(1 point) How much money do winners go home with from the television quiz show Jeopardy? To determine an answer, a random sample of winners was drawn from a normal population and the amount of money each won was recorded and listed below. Estimate with 98% confidence the mean winnings for all the show's players. 25547 32968 32743 24652 27897 23103 29912 31633 19954 27255 17318 26473 16209 28864 24000 UCL = LCL = Note in this problem and future problems, UCL stands for upper confidence level and LCL stands for lower confidence level. These are just the upper and lower bounds of your confidence interval.Explanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =25901.87
standard deviation, s =5187.352
sample size, n =15
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5187.352/ sqrt ( 15) )
= 1339.369
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 14 d.f is 2.624
margin of error = 2.624 * 1339.369
= 3514.503
III.
CI = x ± margin of error
confidence interval = [ 25901.87 ± 3514.503 ]
= [ 22387.367 , 29416.373 ]
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DIRECT METHOD
given that,
sample mean, x =25901.87
standard deviation, s =5187.352
sample size, n =15
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 14 d.f is 2.624
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 = [ 25901.87 ± Z a/2 ( 5187.352/ Sqrt ( 15) ]
= [ 25901.87-(2.624 * 1339.369) , 25901.87+(2.624 * 1339.369) ]
= [ 22387.367 , 29416.373 ]
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interpretations:
1) we are 98% sure that the interval [ 22387.367 , 29416.373 ] 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
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