z scores.

Make sure you get the probability for LEFT outter portion of tail. For example 2.5% on a z value of 1.96

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.

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