statistics help...please
posted by tanya on .
i have no idea!
The real estate assessor for a country government wants to study various characteristics of single family houses in the country. A random sample of 70 houses reveals the following: the mean is 1759, standard deviation is 380 and 42 houses have central air conditioning.
How would I construct a 99% confidence interval estimate of the population mean heated area of the houses?
And the construct a 95% confidence interval estimate of the population proportion of houses that have central air conditioning.

Formula for 99% interval estimate of the population mean:
CI99 = mean + or  2.575(sd/√n)
...where + or  2.575 represents the 99% confidence interval using a ztable, sd = standard deviation, √ = square root, and n = sample size.
Substitute what you know into the formula:
CI99 = 1759 + or  2.575(380/√70)
Finish the calculation.
Formula for 95% interval estimate of the population proportion:
CI95 = p + or  1.96[√(pq/n]
...where + or  1.96 represents the 95% confidence interval using a ztable, p = x/n, q = 1  p, and n = sample size.
Substitute what you know into the formula:
CI95 = 42/70 + or  1.96[√(42/70)(28/70)/70]
Convert the fractions to decimals and finish the calculation.
I hope this will help.