Statistics Question. Need Help!?

[custenhin]

Healthy adult bone marrow containers about 56.5% neutrophils. However, if this level is significantly reduced, it may be an early indicator of leukemia.
a) I have.

b) Suppose Jan had a bone marrow biopsy and p hat was observed to be 0.53. Assuming nothing is wrong (no Leukemia), what is the probability of getting a biopsy result this low or lower? Compute P(p hat equal or less than 0.53).

c) Suppose Meredith had a bone marrow biopsy and p hat was observed to be .41. Assuming nothing is wrong ( no leukemia), what is the probability of getting a biopsy result this low or lower? Compute P(p hat equal or less than 0.41).

Any help would be great!! Thanks in advance!

[bedyw]

Healthy adult bone marrow containers about 56.5% neutrophils. However, if this level is significantly reduced, it may be an early indicator of leukemia.
a) I have.

b) Suppose Jan had a bone marrow biopsy and p hat was observed to be 0.53. Assuming nothing is wrong (no Leukemia), what is the probability of getting a biopsy result this low or lower? Compute P(p hat equal or less than 0.53).

c) Suppose Meredith had a bone marrow biopsy and p hat was observed to be .41. Assuming nothing is wrong ( no leukemia), what is the probability of getting a biopsy result this low or lower? Compute P(p hat equal or less than 0.41).

Any help would be great!! Thanks in advance!

Sorry it took me a while to give you the answer, but I was kinda busy yesterday! Here we go:

Now that we have a sample size of n=50 and a probability of p=0.565 we have a binomial distribution which we can be approximated by the normal distribution (as suggest in a) or according to the central limit theorem). For b) we can then calculate the z-score and look up the probability in a table:

z= (x-µ) / (SQRT(p*(1-p)/N)) ; SQRT = square root

(we use the standard error of the sample instead of the standard deviation)

z= (0.53-0.565)/(SQRT(0.565*0.435/50)= -0.5

Look up in table and you get P(x<0.53) = 40.85%

You do the same thing with in c):

z= (0.41-0.565)/(SQRT(0.565*0.435/50)= -2.21

Look up in table and you get P(x<0.41) = 1.35%

It seems like you have the right answer, so please write me an email if my finding are incorrect. (small rounding affects are possible)

Cheers!

[leon]

Hi -
It's 8 months later... and I'm doing this same question so I'm very happy to have found it online. What I believe accounts for the difference between the correct answer and the answers found in your break down above is that there is no continuity correction being performed.