Question
Asked Nov 4, 2019

In a sample of 30 participants, a researcher estimates the 95% CI for a sample with a mean of Μ12 = 0.7 and an estimated standard error for the difference sΜ1-Μ2 of 0.2. What is the confidence interval at this level of confidence?

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Step 1

From the given question

Sample size (n) =30, mean difference (M1-M2) =0.7 and standard error difference (SM1-M2) =0.2.

MD 0.7, SMD-0.2 and n=30
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MD 0.7, SMD-0.2 and n=30

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Step 2

Formula for confidence interval (CI) is

CI=MD(txSMD) >(1)
Here, t-value is for 95% confidence level and sample size 30
a 1-0.95
=0.025 and degree of freedom (df)=n-1=30-1=29
2
t-value for
2
from the t-distribution table,t
2.045
(0.025,29)
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CI=MD(txSMD) >(1) Here, t-value is for 95% confidence level and sample size 30 a 1-0.95 =0.025 and degree of freedom (df)=n-1=30-1=29 2 t-value for 2 from the t-distribution table,t 2.045 (0.025,29)

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Step 3

Let substitute these values in equation...

CI-MD (txSMD)
0.7+(2.045x0.2
(0.291, 1.109)
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CI-MD (txSMD) 0.7+(2.045x0.2 (0.291, 1.109)

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