Question

1.A random sample of size 120 is drawn from a large population with mean 38.75 obtain the sd of the distribution of all possible sample mean ( let the sample sd be s= 5.28 ) what is the sampling distribution of the mean?

2. In a random sample of size 506, the average cholesterol level of group of adults is 96.997 if the standard deviation of the colesterol level in the city adults population is 1.74 find the 72% confidence for u

Answer 1

Solution:

Question 1)

Given:
Sample size = n = 120

Sample Mean =

Sample Standard Deviation = s = 5.28

the sampling distribution of the mean:

Since sample size n =120 is large , we can use Central limit theorem which states that for large sample size n ,
sampling distribution of sample mean is approximately normal with mean of sample means:

and standard deviation of sample means is:

Since is unknown , we use its sample estimate s = 5.28

Thus

Question 2)

Given:

Sample size = n = 506

Sample Mean =

Population Standard Deviation = = 1.74

We have to find 2% confidence for mean:

where

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.72) /2 = 1.72 / 2 = 0.8600

Look in z table for Area = 0.8600 or its closest area and find z value.

Area = 0.8599 is closest to 0.8600 and it corresponds to 1.0 and 0.08 , thus z critical value = 1.08

That is : Z_c = 1.08

Thus

Thus