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Answer:
The calculated value of t= -0.9468 does not lie in the critical region t= -1.701. Therefore we accept our null hypothesis that mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software at the 0.05 significance level
Step-by-step explanation:
Categories Sample Size Mean Standard D
Traditional Lab 17 71.5 4.8
Software 13 78. 5 6
We are using Student's t- test considering that the population variance is equal.
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≤ -t (0.05, 28) = -1.701
Degrees of freedom is calculated df = υ= n1+n2- 2= 17+13-2= 28
Here the difference between the sample means is x`1- x`2= 71.5-78.5= -7
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/17+13-2 [ 71.5²+78.5²]
Sp = 20.0664
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= -7/ 20.0664 √1/17+ 1/13
t= -7/7.393
t= -0.9468
The calculated value of t= -0.9468 does not lie in the critical region t= -1.701. Therefore we accept our null hypothesis that mean science test score is lower for students taught in traditional lab sessions than it is for students taught using interactive simulation software at the 0.05 significance level