My Notes Ask Your Teac 7 points MintroStat6 7.E.031. In a study of children with

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My Notes Ask Your Teac 7 points MintroStat6 7.E.031. In a study of children with a particular disorder, parents were asked to rate their child on a variety of items related to how well their child performs different tasks. One item was "Has difficulty organizing work," rated on a five-point scale of 0 to 4 with 0 corresponding to "not at all" and 4 corresponding to "very much." The mean rating for 275 boys with the disorder was reported as 2.29 with a standard deviation of 1.09. (Round your answers to four decimal places.) Compute the 90% confidence interval Compute the 95% confidence interval. CL Compute the 99% confidence interval Explain the effect of the confidence level on the width of the interval. We see that the width of the interval decreases with confidence level O We see that the width of the interval does not change with confidence level. O We see that the width of the interval increases with confidence level Submit Assignment Home My Assignments Extension Reques WebAssign®4.0 1997-2017 Advanced Instructional Systems, Inc. Al nghts reserved. Advanced Instructional Systems, nc. Allr Type here to search ^4x 10/18/2017

Explanation / Answer

a)
given that,
sample mean, x =2.29
standard deviation, s =1.09
sample size, n =275
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 274 d.f is 1.65
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 = [ 2.29 ± Z a/2 ( 1.09/ Sqrt ( 275) ]
= [ 2.29-(1.65 * 0.066) , 2.29+(1.65 * 0.066) ]
= [ 2.182 , 2.398 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 2.182 , 2.398 ] 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
b)
given that,
sample mean, x =2.29
standard deviation, s =1.09
sample size, n =275
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 274 d.f is 1.969
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 = [ 2.29 ± Z a/2 ( 1.09/ Sqrt ( 275) ]
= [ 2.29-(1.969 * 0.066) , 2.29+(1.969 * 0.066) ]
= [ 2.161 , 2.419 ]
interpretations:
1) we are 95% sure that the interval [ 2.161 , 2.419 ] 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
c)
given that,
sample mean, x =2.29
standard deviation, s =1.09
sample size, n =275
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 274 d.f is 2.594
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 = [ 2.29 ± Z a/2 ( 1.09/ Sqrt ( 275) ]
= [ 2.29-(2.594 * 0.066) , 2.29+(2.594 * 0.066) ]
= [ 2.119 , 2.461 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 2.119 , 2.461 ] 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

d)
width of the interval increse with the increase confidence interval

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