Colin put some buttons on a table. There were 4 blue buttons, 5 red buttons, 7 tan buttons, and 8 white buttons. Colin's cat jum
2 answers:
To find a probability of an event occurring, we need to find how many outcomes are successful. Then divide that number over the total number of outcomes.
In this case, we have the cat knocking 1 button onto floor. We want that button to be blue. Because we have 4 blue buttons, we have 4 different scenarios/outcomes that are successful.
Total number of outcomes is the cat knocking any button onto the floor. Which is [/tex] 4+5+7+8=24[/tex].
When we divide the two:
is the probability of cat knocking a blue button onto the floor.
In this case, we have the cat knocking 1 button onto floor. We want that button to be blue. Because we have 4 blue buttons, we have 4 different scenarios/outcomes that are successful.
Total number of outcomes is the cat knocking any button onto the floor. Which is [/tex] 4+5+7+8=24[/tex].
When we divide the two:
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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.
