A bank manager wants to know the mean amount of mortgage paid per month by homeo

Question

A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of $1575 per month for their mortgages. The population standard deviation of such mortgages is $215.
Determine:
a) parameter to estimate:
b) estimator :
c) point estimate:
d) What is the sampling distribution of your estimator?
e) Find a 95% Confidence Interval (CI) for the mean amount of mortgage paid per month by all homeowners in this area
f) f)What is the margin of error of estimate from part e
g) Is the true population mean included in this interval?
h) Is the sample mean included in this interval?
I) Is this statement true: Probability (the true population mean is included in this interval above) =0.95. If not, please correct.
j) What does the phase “95% confident “mean?
j) Find a 97% Confidence Interval (CI) for the mean amount of mortgage paid per month by all homeowners in this area
k) Suppose the confidence interval is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

Explanation / Answer

a. Parameter is a numerical characteristic of a population, as distinct from a statistic of a sample . So parameter here is the mean amount of mortgage paid per month by homeowners in an area.  

b. Estimator is a quantity used or evaluated as an estimate of the value of a parameter. Here estimator is an average of $1575 per month for their mortgages.

c. Point estimate of mean=mu=1575

d. As per central limit theorem when n is sufficient large sampling distribution of mean is normal with mean=mu=1575 and sd=sd/sqrt(n)=*215/sqrt(120)=19.63

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.