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Answer:
It is concluded that no difference exists in the comprehension of the lab based on the test scores.
Step-by-step explanation:
<em><u>Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.</u></em>
The above statement tells us that the independent t or two sample t test will be performed as both have equal variances and are normally distributed.
This can be easily done through excel.
The following table is obtained
t-Test: Two-Sample Assuming Equal Variances
<u> Text Visual Illustrations</u>
Mean 70.28 75.08
Variance 304.48 228.58
Observations 15 15
Pooled Variance Sp²= 266.53
Pooled Standard Deviation = Sp = 16.33
Hypothesized Mean Difference = x1`-x2`= 0
df = n1+n2-2= 15+15-2= 30-2= 28
t Stat -0.805188239
<u>P(T<=t) two-tail 0.427495979 </u>
<u>t Critical two-tail 1.701130908 </u>
Let the null and alternate hypotheses be
H0 : u1-u2= 0 against the claim Ha: u1-u2≠0
There is no difference between the means
against the claim
that there is a difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.
The significance level is ∝= 0.1
The d.f is n1+n2-2= 15+15-2= 30-2=28
This is a two tailed test and the critical region is t (0.025) (28) ≥ 1.7011 and t (0.025) (28) ≤ - 1.7011.
The test statistic is
t= x1-x2/ Sp √1/n1+ 1/n2
t= 70.28 -75.08/ 16.33√1/15 +1/15
t= -4.8/5.963
t= -0.8049811 ( minute difference from excel result due to rounding)
Since the calculated value of t= -0.8049 does not fall in the critical region we conclude that there is no difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.
We accept the null hypothesis.
The p- value is 0.427495979.
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)