In a study of academic procrastination, the authors of a paper reported that for
ID: 3319557 • Letter: I
Question
In a study of academic procrastination, the authors of a paper reported that for a sample of 441 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.34 hours and the standard deviation of study times was 3.50 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university. (a) Construct a 95% confidence interval to estimate , the mean time spent studying for the final exam for students taking introductory psychology at this university. (Round your answers to three decimal places.) (b) The paper also gav the exam. e the following sample statistics for the percentage of study time that occurred in the 24 hours prior to n=441 x=43.48 s=21.46 confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam. Construct a 90% (Round your answers to three decimal places.)Explanation / Answer
a.
TRADITIONAL METHOD
given that,
sample mean, x =7.34
standard deviation, s =3.5
sample size, n =441
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.5/ sqrt ( 441) )
= 0.167
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 = 440 d.f is 1.965
margin of error = 1.965 * 0.167
= 0.328
III.
CI = x ± margin of error
confidence interval = [ 7.34 ± 0.328 ]
= [ 7.013 , 7.668 ]
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DIRECT METHOD
given that,
sample mean, x =7.34
standard deviation, s =3.5
sample size, n =441
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 440 d.f is 1.965
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 = [ 7.34 ± t a/2 ( 3.5/ Sqrt ( 441) ]
= [ 7.34-(1.965 * 0.167) , 7.34+(1.965 * 0.167) ]
= [ 7.013 , 7.668 ]
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interpretations:
1) we are 95% sure that the interval [ 7.013 , 7.668 ] 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
b.
TRADITIONAL METHOD
given that,
sample mean, x =43.48
standard deviation, s =21.46
sample size, n =441
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 21.46/ sqrt ( 441) )
= 1.022
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 440 d.f is 1.648
margin of error = 1.648 * 1.022
= 1.684
III.
CI = x ± margin of error
confidence interval = [ 43.48 ± 1.684 ]
= [ 41.796 , 45.164 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =43.48
standard deviation, s =21.46
sample size, n =441
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 440 d.f is 1.648
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 = [ 43.48 ± t a/2 ( 21.46/ Sqrt ( 441) ]
= [ 43.48-(1.648 * 1.022) , 43.48+(1.648 * 1.022) ]
= [ 41.796 , 45.164 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 41.796 , 45.164 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean