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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 . |
Your review questions and global ranking system were so helpful.
Lina
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® 2024 Level I Curriculum, Volume 1, Module 8.