A randomly selected sample of n 12 students at a university is asked, \"How much

Question

A randomly selected sample of n 12 students at a university is asked, "How much did you spend for textbooks this semester?' The responses, in dollars, are the 200 175 450 300 350 250 150 200 320 370 404 250 Computer output with a 95% confidence interval for the population mean is shown in the following figure. Use the Empirical Rule to create an interval that estimates the textbook expenses of 95% of the individual students at the university. (Round the answer to the nearest dollar) to $ variable | N | Mean | StDev | SEMean | 95.0% ci spent 12 284.9 96.1 27.7 (223.9,345.0 You may need to use the appropriate table in the Appendix of Tables to answer this question. Noed Help? Talk to ter

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =284.9167
standard deviation, s =96.0639
sample size, n =12
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 96.0639/ sqrt ( 12) )
= 27.7
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 2.201
margin of error = 2.201 * 27.7
= 61
III.
CI = x ± margin of error
confidence interval = [ 284.9167 ± 61 ]
= [ 223.9 , 346 ]
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DIRECT METHOD
given that,
sample mean, x =284.9167
standard deviation, s =96.0639
sample size, n =12
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 2.201
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 = [ 284.9167 ± t a/2 ( 96.0639/ Sqrt ( 12) ]
= [ 284.9167-(2.201 * 27.7) , 284.9167+(2.201 * 27.7) ]
= [ 223.9 , 346 ]$
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 223.9 , 346 ] 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

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