Your statistics professor assi
Your statistics professor assigns nice tests, cruel tests ordiabolical tests. Nice tests are the easiest, diabolical tests arethe hardest, and cruel tests are in between. On cruel tests, scoresare normally distributed with μc= 52 and σc= 14. Nice anddiabolical tests have different μ and different σ. Your class has49 students. The average score for the class of 49 students isXbar.
(a) If the first test is cruel, what is the probability that Xbar is between 54 ~ 58?
(b) The average score on the first test is 47. Test the nullhypothesis that the first test was cruel. Use α = .05 and atwo-sided test to test this null hypothesis.
(c) Suppose that nice tests also have scores that are normallydistributed, but they have a different μ and σ, with μn = 55 and σn= 10.5. For the second test of the semester, you plan to do anotherhypothesis test, again with the null hypothesis that the test iscruel. If the second test is in fact nice, not cruel, what is theprobability you will make a type 2 error in your hypothesistest?
(d) The professor says that the final test will have a 50%chance of being cruel and a 50% chance of being nice. For thistest, the professor does not report the class average, instead herandomly picks our one exam and tells you the score. If therandomly chosen exam is above an 80, what is the probability it wasnice?
Answer:
(a)
The z-score for 
is
The z-score for 
is
The probability that X bar is between 54 and 58 is
(b)
(c)
(d)
If the test is cruel then the probability that score is above 80is:
The z-score for C = 80 is
So,
P(Score above 80 | cruel) = P(z > 2) = 0.0228
If the test is nice then the probability that score is above 80is:
The z-score for N = 80 is
So,
P(Score above 80 | nice) = P(z > 2.38) = 0.0087
And we have
P(cruel) = P(nice) = 0.50
By the Baye’s theorem we have
P(nice | score above 80) = [ P(Score above 80 | nice)P(nice) ] /[ P(Score above 80 | nice)P(nice) + [ P(Score above 80 |cruel)P(cruel) ] ] = [ 0.0087 *0.5 ] / [ 0.0087 *0.5 + 0.0228 * 0.5] = 0.2762
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