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Basic Question 12 of 13
A floor manager of a large department store is studying the buying habits of the store's customers. Suppose he assumes that the monthly income of these customers is normally distributed with a standard deviation of 500. If he draws a random sample of size N = 100 and obtains a sample mean of YBAR = 800, a 0.95 confidence interval for the true population mean is 702 < MU < 898. Do you think that it would be quite unreasonable for the true population mean to be $600? Explain.
User Contributed Comments 4
| User | Comment |
|---|---|
| topdan | Any more detailed explanation please ? |
| ehc0791 | how about this 1. H(o): x <= 600 vs H(a): x > 600 2. rejection point 1.96 3. if z > 1.96, reject H(o) 4. z = (800 - 600)/(500/sqrt(100)) = 4 |
| mchu | It is in the notes. Confidence intervals can be used to test hypotheses. Note that the right side of the equation is the left endpoint of the confidence interval. Essentially, if the confidence interval contains the value of the unknown population parameter as hypothesized under H0, then H0 would not be rejected in a two-sided hypothesis test with corresponding α, whereas if the confidence interval does not contain the value of the unknown population parameter as hypothesized under H0, then H0 would be rejected in a two-sided hypothesis test with corresponding α. Since 600 is not in the confidence interval, the H0 would be rejected. |
| chamad | Plese note this is 2 tailed rest H0 : Income = 600 Ha : income != 600 The confidence intervl is (702 < MU < 898) So obviously 600 is outside this interval Reject H0 |
I used your notes and passed ... highly recommended!
Lauren
Learning Outcome Statements
explain hypothesis testing and its components, including statistical significance, Type I and Type II errors, and the power of a test
CFA® 2024 Level I Curriculum, Volume 1, Module 8.