Paired sample t-test is used.
For the difference in scores After-Before:
"n=10,\\;\\bar{x}=1,\\;s=4.27."
Null hypothesis "H_0:\\;\\mu_d=0."
Alternative hypothesis "H_a:\\;\\mu_d>0."
Test statistic: "t=\\frac{\\bar{x}}{\\frac{s}{\\sqrt{n}}}=\\frac{1}{\\frac{4.27}{\\sqrt{10}}}=0.74."
Degrees of freedom: "df=n-1=10-1=9."
P-value: "p=P(T>0.74)=0.2391."
Since P-value is greater than 0.05, fail to reject the null hypothesis.
There is no sufficient evidence that the employees benefited by the training.
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