statistics help...please

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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.

  • statistics help...please -

    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 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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