Question

How many children would be required to ensure that the margin of error in (a) does not exceed 10 units?

2. Consider again the study in Question 1, a different investigator conducts a second study to investigate whether there is a difference in mean PEF in children with chronic bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at D=0.05. (20 points, 4 points each) Group Number of Children|Mean PEFSud Dev PEF Chronic Bronchitis 25 281 No Chronic Bronchitis 25 68 319 1) Write down the hypothesis 2) What is the formula of test statistics? 3) What is the critical value? What is the test value? (show your work below) s» What is the conclusion? What does the result mean? 3. Using the data presented in Question 2, Construct a 95% confidence interval for the mean PEF in children without chronic bronchitis (12 points) 1) What is the formula of test statistics? 2) What is the upper limit and lower limit? (show your work below) b) How many children would be required to ensure that the margin of error in (a) does not exceed 10 units? (8 points) 1) What is the formula of test statistics? 2) What is the number of subjects needed? (show your work below)

Answer 1

2.
Given that,
mean(x)=281
standard deviation , s.d1=68
number(n1)=25
y(mean)=319
standard deviation, s.d2 =74
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.492
since our test is left-tailed
reject Ho, if to < -2.492
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =281-319/sqrt((4624/25)+(5476/25))
to =-1.891
| to | =1.891
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 2.492
we got |to| = 1.89057 & | t α | = 2.492
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:left tail - Ha : ( p < -1.8906 ) = 0.03541
hence value of p0.01 < 0.03541,here we do not reject Ho
ANSWERS
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1.
null, Ho: u1 = u2
alternate, H1: u1 < u2
2.
test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
4.
test statistic: -1.891
3.
critical value: -2.492
decision: do not reject Ho
p-value: 0.03541
5.
we do not have enough evidence to support the claim that lower mean PEF in children with chronic bronchitis as compared to those without
3.
a.
1.
TRADITIONAL METHOD
given that,
sample mean, x =319
standard deviation, s =74
sample size, n =25
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 74/ sqrt ( 25) )
= 14.8
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.064
margin of error = 2.064 * 14.8
= 30.547
III.
CI = x ± margin of error
confidence interval = [ 319 ± 30.547 ]
= [ 288.453 , 349.547 ]
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2.
DIRECT METHOD
given that,
sample mean, x =319
standard deviation, s =74
sample size, n =25
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 24 d.f is 2.064
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 = [ 319 ± t a/2 ( 74/ Sqrt ( 25) ]
= [ 319-(2.064 * 14.8) , 319+(2.064 * 14.8) ]
= [ 288.453 , 349.547 ]
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interpretations:
1) we are 95% sure that the interval [ 288.453 , 349.547 ] 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.
given data,
margin of error =10
standard deviation = 74
confidence level is 95%
1.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 74
ME =10
n = ( 1.96*74/10) ^2
= (145.04/10 ) ^2
= 210.366 ~ 211          
2.
sample size = 211