write the answer on a paper plz Suppose a teaching method st

Suppose a teaching method study was designed to test a hypothesis of equal means on the final examinations scores for an experimental teaching method and the traditional lecture method. Participants were randomly assigned to one of the two methods, classes were taught, and final examination scores were recorded. A summary of the data is as follows. Experiment: n = 16 Sample mean = 87.5 Sample standard deviation = 38.13 Traditional: n = 14 sample mean = 82.0 sample standard deviation = 42.53. Find the 98 percentage interval for the difference of mean score of two methods.

Solution

a)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=87.5
Standard deviation( sd1 )=38.13
Sample Size(n1)=16
Mean(x2)=82
Standard deviation( sd2 )=42.53
Sample Size(n1)=14
CI = [ ( 87.5-82) ±t a/2 * Sqrt( 1453.8969/16+1808.8009/14)]
= [ (5.5) ± t a/2 * Sqrt( 220.07) ]
= [ (5.5) ± 2.65 * Sqrt( 220.07) ]
= [-33.81 , 44.81]

b)
Set Up Hypothesis
Null , There Is No-Significance between them Ho: u1 = u2
Alternate, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=87.5
Standard Deviation(s.d1)=38.13 ; Number(n1)=16
Y(Mean)=82
Standard Deviation(s.d2)=42.53; Number(n2)=14
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =87.5-82/Sqrt((1453.8969/16)+(1808.8009/14))
to =0.371
| to | =0.371
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 13 d.f is 2.16
We got |to| = 0.37075 & | t | = 2.16
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 0.3708 ) = 0.717
Hence Value of P0.05 < 0.717,Here We Do not Reject Ho

Two methods would n\'t make any diffrence in average score of students