Question

1) If you are a human resources manager and you randomly selected 5 job candidates to interview out of hundreds of equally-qualified candidates that applied for a job without replacement, what would be the standard error of the mean if the standard deviation is 1?

2) The U.S. Census Bureau announced that the population mean sales of houses was $322,100. Assume that the standard deviation of the prices is $90,000. If you select a sample size of 100, what’s the cumulative probability that the sample mean will be less than $346,000 for all possible house samples? (Tip: First find the Z value, then find the cumulative probability to determine the percentage)

What’s the cumulative probability according to the Z-table?

3) A survey manager wishes to find out the sample size that is needed at a particular confidence level. If the manager finds out that the ? = 20, what sample size is needed to estimate the mean within +/-5 with 95% (1.96) confidence?

Answer 1

Anas:

2)Given that

population mean=322100

standard deviation=90000

sample mean=346000

n=100

standard error of mean=90000/sqrt(100)=9000

z=(346000-322100)/9000

z=2.66

P(z<2.66)=0.9960

3)Given that

Marginof error=5

z=1.96

Sample size required,n=(1.96*20/5)^2=61.47

1)standard error of mean=s/sqrt(n)=1/sqrt(5)=0.45