Question

Part 1: The amount of time that users of a real estate website spend on the site is believed to be normally distributed. Suppose 20 randomly-selected users of the site were asked how much time they typically spend on the site during the week, and the sample mean was found to be 5.2 hours.

1. (5 pts.) If the population standard deviation is known to be σ=2.8 hours, then
   
   1. What is the point estimate for µ in this problem?

2. Find the margin of error for a 99% confidence interval for µ. Round to the nearest tenth.

3. Find the 99% confidence interval for µ.

2. (5 pts) If the population standard deviation is unknown, but the sample standard deviation is
   s=2.89 , then
   
   1. What is the point estimate for µ in this problem?

2. Find the margin of error for a 99% confidence interval for µ. Round to the nearest tenth.

3. Find the 99% confidence interval for µ.

Answer 1

a) The point estimate for the population mean is equal to the sample mean which is given here to be 5.2 hours. therefore 5.2 hours is the point estimate here.

b) From standard normal tables, we have:
P( -2.576 < Z < 2.576) = 0.99

Therefore the margin of error here is computed as:

Therefore, 1.6128 is the required margin of error here.

c) The 99% confidence interval here is obtained as:

This is the required 99% confidence interval for population mean here.

d) The points estimate for the population mean still remains the same which is 5.2 hours here.

Therefore 5.2 hours is the point estimate value here.

e) For n - 1 = 19 degrees of freedom, we have from t distribution tables:

P( t₁₉ < 2.861) = 0.995

Therefore, due to symmetry, we have here:
P( -2.861 < t₁₉ < 2.861) = 0.99

Now the margin of error here is computed as:

Therefore 1.8488 is the required margin of error here.

f) The confidence interval for population mean here is obtained as:

This is the required confidence interval for the population mean here.