solution
part a)
To compare means between 2 samples, we obtain the t value as:
Mean of test, T1:
Variance of test, T1:
Mean of test, T2:
Variance of test, T2:
Therefore, t value:
"t\\ value =0.083449"
The corresponding probability is "0.5324" which is greater than the p value, 0.05
"Degrees\\ of \\ freedom, df = min(df_1,df_2) =10"
The critical value at "\\alpha =0.05"
We conclude that there is no benefit from the coaching since the t value calculated (0.083449) is less than the critical value (2.228) and the probability (0.5324) is greater than 0.05
part b)
The t statistic given "\\mu=47.5"
"t \\ value = 2.039974"
The corresponding probability is "0.102223" which is greater than the p value 0.05
"Degrees\\ of \\ freedom, df=n-1=8"
The critical value at a 95% level of confidence
Since the t value (2.039974) is less than the critical value (2.306), and the p value (0.102223) is greater than 0.05, we conclude that the mean of the 9 items does not differ significantly from the assumed population with mean 47.5
Comments
In Part a) the corresponding probability is given by P(Y>0.083449)=0.5324, where Y has the t-distribution with degrees of freedom df=10. In Part b) the corresponding probability is given by P(T2.039974)=2P(T>2.039974)=0.102223, where T has the t-distribution with degrees of freedom df=8.
How can I find out corresponding probability in above mentioned questions
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