Your z-table gives, assuming your z-scores are correct:Originally Posted by Jimmy23
pr(z<-1.79)=0.0367,
and:
pr(z<2.68)=0.9963.
Therefore:
pr(-1.79<z<2.68)=0.9963-0.0367=0.9596
RonL
Well I thought I was doing this right until I got to a certain part of the question. Here is the problem: The National Health and Nutrition Examination Survey of 1976-1980 found that the mean serum cholesterol level for US males aged 20 to 24 years was 180. The standard deviation was approximately 43.
a) Consider the sampling distribution of the sample mean based on random samples of size 60 drawn from this population of males. What is the mean and standard error of this sampling distribution?
So here I said the mean remaind the same, so 180 and the standard error would be 43/sqr rt of 60 so 43/7.7 = 5.6
b) If a random sample of size 60 is drawn from the described population,
1) find the probability that the sample mean cholesterol level will be below 170 or above 195.
Here's where I'm confused. Doing my z-score formula I get a z-score of -1.79 for 170 and 2.68 for 195. I take those numbers to my z-table and get .0367 and .9963. Then I added those two values together and get 1.033 and subtract that value from 1, which = -0.033. So is this all completely wrong?
Part 2 of the question, which I think I got right, but confirmation would be nice
What is the probability that the sample will yield a mean cholesterol level greater than 175?
so z=175-180/43/sqr rt of 60 = -5/5.6=-.89 I go to my table and get .1867 so 1-.1867 = .8133
Once again, thanks for all the help!
The short answer is no.Originally Posted by Jimmy23
The questions given ask about the sampling distribution of the sample mean when the population standard deviation is known. Then the sample mean will have a known normal distribution, so you can translate to a z-score and use a normal distribution table to answer the questions as you did.
The t-test uses a t-statistic and a t-distribution table. The t-statistic is used when the population standard deviation is unknown and must be estimated from the sample. The t-statistic is like the z-score but it has the sample mean in the numerator and the sample standard deviation in the denominator. The t-distribution gives the sampling distribution of the t-statistic.
The questions given did not say anything about the sample standard deviation or the t-statistic. So the t-test does not apply.