Question

In: Statistics and Probability

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of...

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 225 students using Method 1 produces a testing average of 68.2 A sample of 242 students using Method 2 produces a testing average of 66.2. Assume that the population standard deviation for Method 1 is 5.66 while the population standard deviation for Method 2 is 10.06. Determine the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 3 of 3 :  

Construct the 98% confidence interval. Round your answers to one decimal place.

Solutions

Expert Solution

TRADITIONAL METHOD
given that,
mean(x)=68.2
standard deviation , σ1 =5.66
population size(n1)=225
y(mean)=66.2
standard deviation, σ2 =10.06
population size(n2)=242
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((32.0356/225)+(101.2036/242))
= 0.7
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 0.7
= 1.7
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (68.2-66.2) ± 1.7 ]
= [0.3 , 3.7]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=68.2
standard deviation , σ1 =5.66
number(n1)=225
y(mean)=66.2
standard deviation, σ2 =10.06
number(n2)=242
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 68.2-66.2) ±Z a/2 * Sqrt( 32.0356/225+101.2036/242)]
= [ (2) ± Z a/2 * Sqrt( 0.6) ]
= [ (2) ± 2.326 * Sqrt( 0.6) ]
= [0.3 , 3.7]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [0.3 , 3.7] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.02 true mean
difference is zero


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