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March 3, 2015

March 3, 2015

Posted by **tanya** on Thursday, April 3, 2008 at 11:23am.

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.

- statistics help...please -
**MathGuru**, Thursday, April 3, 2008 at 12:53pmFormula 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 z-table, 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 z-table, 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.

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