Following is the output of descriptive statistics and box
plot:
![Descriptive statistics Pessimistic, X1 Optimistic, X2 count 10 10 2.90 6.80 mean sample standard deviation 2.13 1.69 4.54 2.8](//img.homeworklib.com/questions/0504b190-2e29-11ea-b6f1-af0c20d1b59e.png?x-oss-process=image/resize,w_560)
Box plots shows that both distributions are approximately
symmetric and there is no outlier so we can assume that data come
from normal population.
That is we can use t test for independent samples
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![Here we have -2.9, 5, -2.13, п,310 х, -6.8,5, %3D1.69, п,310 Hypotheses are: 0 - H a Test is two tailed. Level of significanc](//img.homeworklib.com/questions/056d8280-2e29-11ea-a610-532762700997.png?x-oss-process=image/resize,w_560)
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Conclusion: There is evidence to conclude that recycling
behaviors differed between groups.
Descriptive statistics Pessimistic, X1 Optimistic, X2 count 10 10 2.90 6.80 mean sample standard deviation 2.13 1.69 4.54 2.84 sample variance minimum 0 maximum 6 9 5 range ВоXPlot 0 1 2 4 5 6 7 Pessimistic, X1 ВоХPlot 2 4 5 6 7 8 9 10 Optimistic, X2
Here we have -2.9, 5, -2.13, п,310 х, -6.8,5, %3D1.69, п,310 Hypotheses are: 0 - H a Test is two tailed. Level of significance: a-0.05 Since we can assume that variances are equal so degree of freedom of t test will be: df 3Dn +п, -2 -18 The pooled standard deviation is: (n-1)s(n-1)s2, -1.9226 п+п, -2 Test Statistics: (-(4 ((2.9-6.8)-0)/(1.9226* sqrt((1/10)+(1/10)) t= 1 1 =-4.54 п, Critical value: 2.101 Rejection Region: If Itl >2.101, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value: P-value-0.0003 Decision: Since p-value is less than level of significance so we reject the null hypothesis =TINV (0.05,18) =TDIST(4.54,18,2) Excel function for critical value: Excel function for p-value: