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Basic Question 0 of 9
A consumer group wants to prove that average hospital costs are more than $931 per day. The group randomly samples 60 accounts and finds a sample mean of $950. The hypothesis test is at an α-level of 5% and assumes a σ of $50. The critical value is ______ standard deviations.
B. -1.645
C. 1.28
A. 1.645
B. -1.645
C. 1.28
User Contributed Comments 7
| User | Comment |
|---|---|
| Bibhu | One can refer to www.mathsnet.net/asa2/modules/s33binom.html for further reading on this subject. The easier way is sample mean 950 above 931. So +Z and then 5% means 1.645. |
| whiteknight | we can approach the problem like this...for 5% level of significance in a one tail test the z value is 1.645...i.e the right hand portion of the normal distribution....thus for any Z greater than 1.645 this statement will hold true ... |
| raymondg | could somebodyplease explain how to arrive to 1.645 ? |
| meghanchloe | I THOUGHT 5% LEVEL OF SIGNIFICANT IS 1.96 |
| ebayer | 5% one tail = 10% both tails -> use 1.645 instead of 1.96. |
| johntan1979 | This test is clearly one-tailed (>931). So answer should be 1.645 if alpha is .05. two-tailed would be 1.96. |
| Sp1993 | Hi johntan1979, could you please clarify why a two-tailed test would give an answer of 1.96! Thanks! |
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Learning Outcome Statements
calculate and interpret the predicted value for the dependent variable, and a prediction interval for it, given an estimated linear regression model and a value for the independent variable
CFA® 2026 Level I Curriculum, Volume 1, Module 10.