4-2 1/4 as a fraction !!! ill mark !! have a good day !!
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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)
