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Basic Question 2 of 8
A consumer group wants to prove that the 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 a-level of 5% and assumes a s of $50. The p-value is ______% (to the nearest 0.1%).
B. 0.2%
C. 99.8%
A. 5.0%
B. 0.2%
C. 99.8%
User Contributed Comments 13
| User | Comment |
|---|---|
| GeoffT | Need z-table |
| danlan | (950-931)/(50/sqrt(60))=2.94, how to get 0.9984? |
| lockedin | From the z-table, half the area under the curve for 2.94 = 0.4984. The total area for 2.94 is 0.4984 x 2 = 0.9968. Therefore, p = 1 - 0.9968 = 0.32%. I think the 0.9984 from the answer is wrong. Anyone else? |
| charlie | The answer is correct. Notice this is a one-tailed test so 0.5 + 0.4984 = 0.9984. |
| Janey | YOu dont need to calculate this question. P is always less than the level of significance so answer has to be 0.2% |
| xcye | Janey, that's not true, p-value is only less than the level of significance if you reject the Ho! |
| SuperKnight | In this case you don't really need to calculate the p-value to know that it is lower than 5%, when you get the z-value of 2.94.. you know that its greater than 2.58 which makes up the 99% confidence interval (meaning alpha is 0.5% in each tail).. so you know that your value has to be even lower than 0.5%, in this case being 0.2%. |
| bantoo | great wisdom and common sense janey |
| johntan1979 | SuperKnight is THE MAN! |
| gill15 | Janey: P is not always less then significance. There can be large p values and that would indicate we would NOT reject the null hypothesis. Thats the point of calculating it. |
| sgossett86 | I get how we found the Z score, and I get that it gets plugged into the equation and that it will come out as a probability using (1-x), but I don't really understand anything else about it. It doesn't seem like we used Alpha at all in the problem.. and what is this answer really saying?? |
| Shaan23 | Sgoss - we are NOT using alpha in these questions. First all we're doing is finding the z critical value which is 2.94 here. That percentage when you look up in the Ztable is .2%. That .2% all it means is that value of significance is REQUIRED for the null hypothesis to be rejected. |
| jgoff508 | Wow, helpful question . |
I am using your study notes and I know of at least 5 other friends of mine who used it and passed the exam last Dec. Keep up your great work!
Barnes
Learning Outcome Statements
construct hypothesis tests and determine their statistical significance, the associated Type I and Type II errors, and power of the test given a significance level
CFA® 2025 Level I Curriculum, Volume 1, Module 8.