CFA Practice Question
There are 985 practice questions for this topic.
CFA Practice Question
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%).
Correct Answer: B
The p-value is P(x-bar = 950 given a population mean of 931 or less). The z-score for 950 based on this assumption is 2.94. The table value for 2.94 is 0.9984. So, p = 1 - 0.9984 = 0.0016 = 0.2% (to the nearest 0.1%).
User Contributed Comments 13
|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 .|