Confidence Intervals and Means
Message posted by Willy on November 25, 2000 at 12:00 AM (ET)

I looked over your pages but could not find any problems such as these. Please help.

The US engergy Information Administration surveys households to obtain data on monthly fuel expenditures for household vehicles. Results of these surveys are contained in Residential transportaion Engergy Consumption Survey, Consumption of Household Vehicles.

Suppose one such random sampling of 63 households found an average monthly fuel expenditure of $58.90, with standard deviation $12.75, estimate the mean monthly fuel expenditure.

A sudy of alternative medicine use by Americans revealed that 39% of the Americans surveyed used at least one unconventional therapy in 1990. The margin of error was given as + or - 2.4 percent. Use this to obtain the confidence interval (a 95% confidence interval)for the percentage of all Americans who, in 1990, wasted their time and risked their health on t hese unproven methods.

Any help or forumlas that you have will be greatfully appreciated because I seem to be stuck on these two problems

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Re: Confidence Intervals and Means
Message posted by nancy diehl on November 30, 2000 at 12:00 AM (ET)

The second question:
The percentage is 39% with margin of error of 2.4 and you want a 95% CI.
The only thing you need is the Z-value associated with a 95% CI, which is 1.96.
The formula is: p +/- Z*M.E.
so 39 +/- 1.96(2.4) ==> 39 +/- 4.7
With 95% confidence, the true percentage of Americans that used at least one
unconventional therapy falls between 34.3% to 43.7%