SOLUTION: My younger brother had a run in earlier with M�decins Sans Fronti�res. He narrowly escaped from an adverse verdict by the court........ What he wants is that he be left alone to

Question 1057876: My younger brother had a run in earlier with M�decins Sans Fronti�res. He narrowly escaped from an adverse verdict by the court........ What he wants is that he be left alone to run his small caf�......
He asked my oldest brother if he can conduct a survey for him about justice in the Canadian Court.
An initial survey was performed right after M�decins Sans Fronti�res accused my brother of wrong doing. Of 1852 customers, 53 were against the aggressive tactics of M�decins Sans Fronti�res. After my brother was cleared by the court, a follow-up survey was performed. Of 4699 customers, 1751 said they did not agree with the aggressive tactics of M�decins Sans Fronti�res.

My question below please show work
At the 1% significance level, do the data suggest that a higher percentage of customers were against M�decins Sans Fronti�res after the court case?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
An initial survey was performed right after M�decins Sans Fronti�res accused my brother of wrong doing. Of 1852 customers, 53 were against the aggressive tactics of M�decins Sans Fronti�res. After my brother was cleared by the court, a follow-up survey was performed. Of 4699 customers, 1751 said they did not agree with the aggressive tactics of M�decins Sans Fronti�res.
My question below please show work
At the 1% significance level, do the data suggest that a higher percentage of customers were against M�decins Sans Fronti�res after the court case?
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Ho: p2 - p1 <= 0
Ha: p2 - p1 > 0 (higher percent were against)
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p2 = 1751/4699 = 0.373
p1 = 53/1852 = 0.029
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sample difference:: p2-p1 = 0.3436
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z(0.3436) = (0.3436-0)/sqrt[(0.373*0.6567/4699) + (0.029*0.971/1852)] = 41.87
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Note: That difference is 41 standard deviations above the mean.
---
Conclusion:: That z-value is so high no p-value can be calculated.
That p-value is so small you could definitely reject Ho and conclude
that tke test results support Ha at the 1% significance level.
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Cheers,
Stan H.
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