13. 5/9 points | Previous Answers DevoreStat9 7.E.037 My Notes A study of the ab

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13. 5/9 points | Previous Answers DevoreStat9 7.E.037 My Notes 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. 0.92 0.85 0.92 0.95 0.93 0.88 1.00 0.92 0.85 0.81 0.75 0.93 0.93 1.05 0.93 1.06 1.09 0.96 0.81 0.97 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 cadence Variable cadence Mean Median TrMean StDe SEMean 20 .9255 0.9300 0.9261 0.0862 0.0193 Min 0.7500 1.0900 0.8650 0.9650 Max 03 (a) Calculate and interpret a 95% confidence interval for population mean cadence. (Round your answers to four decimal places.) 0.8875 , 0.9635 ) strides per second Interpret this interval with 95% confidence, the value of the true mean cadence of all such men falls below the confidence interval. 0 with 95% confidence, the value of the true mean cadence of all such men falls inside the confidence interval. with 95% confidence, the value of the true mean cadence of all such men falls above the confidence interval. (b) Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population. (Round your answers to four decimal places.) 0.9171 , 0.9339 Xstrides per seconc Interpret this interval If this bound is calculated once, there is a 5% chance that these bounds will capture a future individual value of cadence for a healthy man. If this bound is calculated once, there is a 95% chance that these bounds will capture a future individual value of cadence for a healthy man. If this bound is calculated sample after sample, in the long run, 95% of these bounds will capture a future individual value of cadence for a healthy man. 0 If this bound is calculated sample after sample, in the long run, 95% of these bounds will fail to capture a future individual value of cadence for a healthy man. (c) 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.) 0.9145 0.9365 strides per second

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample mean, x =0.9255
standard deviation, s =0.0862
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0862/ sqrt ( 20) )
= 0.0193
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 = 19 d.f is 2.093
margin of error = 2.093 * 0.0193
= 0.0403
III.
CI = x ± margin of error
confidence interval = [ 0.9255 ± 0.0403 ]
= [ 0.8852 , 0.9658 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =0.9255
standard deviation, s =0.0862
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 = [ 0.9255 ± t a/2 ( 0.0862/ Sqrt ( 20) ]
= [ 0.9255-(2.093 * 0.0193) , 0.9255+(2.093 * 0.0193) ]
= [ 0.8852 , 0.9658 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 0.8852 , 0.9658 ] 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.
confidence interval = [ 0.9255 ± 0.0403 ]
= [ 0.8852 , 0.9658 ]
If this bound sample after sample , in long run 95% bound will capture a future indivitual
c.
TRADITIONAL METHOD
given that,
sample mean, x =0.9255
standard deviation, s =0.0862
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0862/ sqrt ( 20) )
= 0.0193
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 2.8609
margin of error = 2.8609 * 0.0193
= 0.0551
III.
CI = x ± margin of error
confidence interval = [ 0.9255 ± 0.0551 ]
= [ 0.8704 , 0.9806 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =0.9255
standard deviation, s =0.0862
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 2.8609
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 = [ 0.9255 ± t a/2 ( 0.0862/ Sqrt ( 20) ]
= [ 0.9255-(2.8609 * 0.0193) , 0.9255+(2.8609 * 0.0193) ]
= [ 0.8704 , 0.9806 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 0.8704 , 0.9806 ] 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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