6. (12 Pts) Mr. Crandall has assigned a term paper due at the end of the semeste

Question

6. (12 Pts) Mr. Crandall has assigned a term paper due at the end of the semester. A random sample of 10 term papers was obtained and the sample mean is 14.7 pages and the sample standard deviation is 5.3 pages. From previous studies, the population distribution of the number of typed pages, is a normal distribution. A confidence interval is to be computed for the population mean number of typed pages of all term papers in Mr. Crandall's class. a) List each assumption and show how each one is met or not met. If the assumptions are met, use StatCrunch to compute a 99% confidence interval for the population mean number of typed pages of all term papers in Mr. Crandall's class b) c) Write the interpretation of the interval in the context of this problem.

Explanation / Answer

6.

TRADITIONAL METHOD
given that,
sample mean, x =14.7
standard deviation, s =5.3
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5.3/ sqrt ( 10) )
= 1.676
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 * 1.676
= 5.447
III.
CI = x ± margin of error
confidence interval = [ 14.7 ± 5.447 ]
= [ 9.253 , 20.147 ]
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DIRECT METHOD
given that,
sample mean, x =14.7
standard deviation, s =5.3
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 = [ 14.7 ± Z a/2 ( 5.3/ Sqrt ( 10) ]
= [ 14.7-(3.25 * 1.676) , 14.7+(3.25 * 1.676) ]
= [ 9.253 , 20.147 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 9.253 , 20.147 ] 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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