What is the volume of the cube in cubic feet? please help
1 answer:
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Answer:
1. Mean square B= 5.32
2. Mean square E= 16.067
3. F= 0.33
4. p-value: 0.28
Step-by-step explanation:
Hello!
You have the information of 3 groups of people.
Group 1
n₁= 20
X[bar]₁= 3.2
S₁²= 14.3
Group 2
n₂= 20
X[bar]₂= 4.2
S₂²= 17.2
Group 3
n₃= 20
X[bar]₃= 7.6
S₃²= 16.7
1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.
Df B= k-1 = 3-1= 2
Sum Square B is:
∑ni(Ÿi - Ÿ..)²
Ÿi= sample mean of sample i ∀ i= 1,2,3
Ÿ..= general mean is the mean that results of all the groups together.
General mean:
Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5
Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64
Mean square B= Sum Square B/Df B= 10.64/2= 5.32
2. The mean square error (MSE) is the estimation of the variance error (σ →
), you have to use the following formula:
Se²=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>
n₁+n₂+n₃-k
Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u> 915.8 </u> = 16.067
20+20+20-3 57
DfE= N-k = 60-3= 57
3. To calculate the value of the statistic you have to divide the MSB by MSE
4. P(F ≤ F) = P(F
≤ 0.33) = 0.28
I hope you have a SUPER day!
Answer:
Step-by-step explanation:
Hello!
The researcher developed a treatment to teach social skills to youth offenders. To test if the treatment is effective in increasing empathy compared to the standard treatment she randomly selected a group of 9 offenders and applied the new treatment and to another group of 9 randomly selected youth offenders, she applied the standard treatment. (Note: the data corresponds to two samples of 9 units each, so I've used those sizes to conduct the test)
At the end of the treatment, she administers BES to measure their empathy levels. Her claim is that the offenders that received the new treatment will have higher BES scores than those who received the standard treatment.
1) Using the records obtained for both groups, she intends to conduct an independent t-test to analyze her claim.
X₁: BES results of a youth offender treated with the new treatment.
X₂: BES results of a youth offender treated with the standard treatment.
H₀: μ₁ = μ₂
H₁: μ₁ ≠ μ₂
α:0.05
test statistic
p-value: 0.7517
The p-value is greater than the significance level so the decision is to not reject the null hypothesis. This means that at a 5% significance level you can conclude that there is no difference between the mean BES scores of the youth offenders treated with the new treatment and the mean BES score of the youth offenders treated with the standard treatment. The new treatment doesn't increase the levels of social empathy of the youth offenders.
I hope this helps
(Box plot in attachment)
