Problem 4. It is generally accepted that Intelligence Quotient (IQ) test scores are normally distributed with a mean of 100 and population standard deviation of 15. You suspect that an online IQ test is giving peoplehigherscoresthantheirtruelQ.Youobtainasimplerandomsampleof100subjects andfind that their mean test score is an IQ of 103. Assume that the population standard deviation does not change among different IQ tests, and that the assumptions for inference are met. What is the power of this test under the alternative μα-105, that is, that the online IQ test truly inflates scores by, on average, 5 points? (Keep the 1% significance level) |What is the probability of Type II Error under this alternative?

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Problem 4. It is generally accepted that Intelligence Quotient (IQ) test scores are
normally distributed with a mean of 100 and population standard deviation of 15.
You suspect that an online IQ test is giving
peoplehigherscoresthantheirtruelQ.Youobtainasimplerandomsampleof100subjects
andfind that their mean test score is an IQ of 103. Assume that the population
standard deviation does not change among different IQ tests, and that the
assumptions for inference are met.
What is the power of this test under the alternative μα-105, that is, that the
online IQ test truly inflates scores by, on average, 5 points? (Keep the 1%
significance level)
|What is the probability of Type II Error under this alternative?
Transcribed Image Text:Problem 4. It is generally accepted that Intelligence Quotient (IQ) test scores are normally distributed with a mean of 100 and population standard deviation of 15. You suspect that an online IQ test is giving peoplehigherscoresthantheirtruelQ.Youobtainasimplerandomsampleof100subjects andfind that their mean test score is an IQ of 103. Assume that the population standard deviation does not change among different IQ tests, and that the assumptions for inference are met. What is the power of this test under the alternative μα-105, that is, that the online IQ test truly inflates scores by, on average, 5 points? (Keep the 1% significance level) |What is the probability of Type II Error under this alternative?
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