A study of the ability of individuals to walk in a straight line reported the ac
ID: 3358438 • Letter: A
Question
A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n = 20 randomly selected healthy men.
0.93 0.85 0.92 0.95 0.93 0.88 1.00 0.92 0.85 0.81
0.79 0.93 0.93 1.05 0.93 1.06 1.06 0.96 0.81 0.99
A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows.
Variable N Mean Median TrMean StDev SEMean
cadence 20 0.9275 0.9300 0.9278 0.0801 0.0179
Variable Min Max Q1 Q3
cadence 0.7900 1.0600 0.8650 0.9750
Calculate an interval that includes at least 99% of the cadences in the population distribution using a confidence level of 95%. (Round your answers to four decimal places.)
Explanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =0.9275
standard deviation, s =0.0801
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0801/ sqrt ( 20) )
= 0.018
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 = 19 d.f is 2.861
margin of error = 2.861 * 0.018
= 0.051
III.
CI = x ± margin of error
confidence interval = [ 0.9275 ± 0.051 ]
= [ 0.876 , 0.979 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =0.9275
standard deviation, s =0.0801
sample size, n =20
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.861
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 = [ 0.9275 ± t a/2 ( 0.0801/ Sqrt ( 20) ]
= [ 0.9275-(2.861 * 0.018) , 0.9275+(2.861 * 0.018) ]
= [ 0.876 , 0.979 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 0.876 , 0.979 ] 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
we are using 95% confidence interval then
DIRECT METHOD
given that,
sample mean, x =0.9275
standard deviation, s =0.0801
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
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 = [ 0.9275 ± t a/2 ( 0.0801/ Sqrt ( 20) ]
= [ 0.9275-(2.093 * 0.018) , 0.9275+(2.093 * 0.018) ]
= [ 0.89 , 0.965 ]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.