Ramashis Biswas | PGID: 61510121 | Section: H | Ramashis_Biswas2015@isb.edu

Answer 1:

Mean: 163862.13

Confidence Interval 99% 95% 90%

Upper Limit 169257.41 167964.69 167304.17

Lower Limit 158466.84 159759.56 160420.08

Margin of Error 5395.285 4102.565 3442.045

The Margin of Error decreases as level of confidence decreases. This is acceptable, as it is

essentially a trade-off between precision and confidence. At 100% confidence, this would

essentially mean the entire range (not precise at all) while at 0% confidence, this would be

essentially a point estimate the mean of a single sample.

For the 99% CI, required n > (t99%, 1046*S/DMOE) 2, or n ~> 32 and hence CLT holds.

Answer 2

The percentile of the population will be an ordered statistic. In order to estimate the same, we

will need more information on the population distribution. Also, since the population is quite

obviously not normal, it would not be possible to make statements about the percentile from

the confidence intervals of the mean.

Answer 3

If we consider that the house prices do not change over time, then each years sample should

represent a simple random sample (assuming that prices have not gone up on resale either,

and houses sold in one year have an equal opportunity of being sold in the next year). Thus

the distribution of the sample averages, as per CLT will be a Normal Distribution with mean

and Variance 2/n.

Answer 4

The sample has 302 houses aged more than 30, i.e. 29% of the sample is old.

From CLT, the 95% and 99% CI for the proportion of old houses will be:

Ramashis Biswas | PGID: 61510121 | Section: H | Ramashis_Biswas2015@isb.edu

95%: 29% + Z95% (p (1-p))/n = [26%, 32%]

99%: 29% + Z99% (p (1-p))/n = [25%, 32%]

This essentially means that the proportion of old houses in the population has a (95/ 99) %

chance of being in the specified intervals.

Or, in (95/ 99)% of random samples will have a proportion in the specified confidence

intervals.