TREHAN> Activities and Due Dates HW8-Confidence Intervals Resources . 9 Hint che
ID: 3312828 • Letter: T
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TREHAN> Activities and Due Dates HW8-Confidence Intervals Resources . 9 Hint che core: 285/1700 6 of 17 > 10 1 7 Leaf Unit= 10 Felisha calculated the mean of her sample to be 773.0 words, and now she wants to calculate the margin of error of a 99% confidence interval for the mean number of words in all books published after 2010. She assumes that the number of words in children's picture books published after 2010 has the same normal distribution and same standard deviation as the general population of all books published since 1950. Compute the margin of error of a 99% confidence interval for the mean number of words in all children's picture books published after 2010. Give your answer precise to one decimal place. wordsExplanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =773
standard deviation, s =17
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 17/ sqrt ( 10) )
= 5.4
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
margin of error = 3.25 * 5.4
= 17.5
III.
CI = x ± margin of error
confidence interval = [ 773 ± 17.5 ]
= [ 755.5 , 790.5 ]
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DIRECT METHOD
given that,
sample mean, x =773
standard deviation, s =17
sample size, n =10
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
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 = [ 773 ± t a/2 ( 17/ Sqrt ( 10) ]
= [ 773-(3.25 * 5.4) , 773+(3.25 * 5.4) ]
= [ 755.5 , 790.5 ]
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interpretations:
1) we are 99% sure that the interval [ 755.5 , 790.5 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
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