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I’ve ran into a few problems that use different formulas to compute the z score.
Some solutions show z = (observation – population mean)/ standard deviation
Some solutions show z = (observation – population mean)/(Std error)
where std error = (std deviation/Sqrt(population size))the following question used the second equation to compute the z score while I used the first equation [(z = observation – population mean)/ std deviation]
How would I know which one to use? they gave significantly different answers.
Please refer to question B. I am fine with question A.
Ex: Assume that the equity risk premium is normally distributed with a population mean of 6% and a population std. deviation of 18%. Over the last four years, equity returns (relative to the risk free rate) have averaged 2.0%. You have a large client who is very upset and claims that results this poor should never occur. Evaluate your client’s concerns.
A. Construct a 95% confidence interval around the population mean for a sample of four year returns.
B. What is the probability of a 2.0 percent or lower average return over a four year period? 


Some solutions show z = (observation – population mean)/(Std error)
where std error = (std deviation/Sqrt(population size))Hint: use this one when you see a sample size in the question. Only use this hint if you’re really strapped on time though. Could throw a curveball obviously on the test.

googs1484 said:Your not testing the hypothesis of a SAMPLE here. 95% confidence interval is 1.96 plus/minus 6 percent. The probability of would be (.02 – .06)/.18. Take that value and look it up on z table.
Oops sry this was wrong but I’m hoping you caught it. Forgot to multiply 1.96 by 18 percent


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