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Sagot :
A null hypothesis can only be rejected if the calculated value is greater than the critical value in this problem.
As the calculated value is less than critical value, null hypothesis cant be rejected.
Null hypothesis could have been rejected if the value would have been greater than -1.533
What is a t-test?
Used in inferential statistics, t test can be used when we need to compare two different sets value to check if their is a significant difference between them.
T tests can be paired and dependent, like measuring the weight of a same patient before and after a diet or they can be unpaired an independent. Unpaired datasets might sometime have the same variance or unequal variance. Different formular are used to calculate the t-tests involving different datasets. To prove a null or alternate hypothesis, the t values must be checked with t- test table.
How to solve the problem using T-test?
From the table;
Pair Group1 Group2
1 10 12
2 8 9
3 11 11
4 8 10
5 9 12
Mean for group 1= 7.4
Mean for group 2= 10.8
S.D for group 1= 4.6
S.D for group 2= 1.36
now to put in the formula for t testing:
t= [tex]\frac{Mean1-Mean2}{\sqrt{\frac{SD1^2}{n1} +\frac{SD2^2}{n2} } }[/tex]= -2.251
df= 5-1 =4(as n1=n2=n)
For a significance of 0.10, the critical value of p is -1.533
As the critical value is greater than the calculated value, null hypothesis that the average difference is less or equal to zero cannot be rejected.
For more information on t-test, visit:https://brainly.com/question/15870238
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