SOLUTION: an auditor of a large oil corporation has taken random sample of monthly credit card charges for 32 individual(non corporate accounts).She found a mean of 481.76.aSSUME THAT THE P

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Question 862198: an auditor of a large oil corporation has taken random sample of monthly credit card charges for 32 individual(non corporate accounts).She found a mean of 481.76.aSSUME THAT THE POPULATION STANDARD DEVIATION IS 48.2 .A 98% CONFIDENCE INTERVAL ESTIMATE FOR THE POPULKATION MEAN OF INDIVIDUAL CREDIT CHARGES WOULD BE
Answer by Theo(13342) About Me  (Show Source):
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the population standard deviation is 48.2

you sample 32 individuals.
the mean of your sample is 48.76

you calculate the standard error of the distribution of sample means as follows:

standard error (se) = standard deviation (sd) of population divided by square root of sample size.

this becomes:

se = 48.2 / sqrt(32) which is equal to 8.5206 rounded to 4 decimal places.

a 98% confidence limit corresponds to a z-score of from -2.33 to + 2.33 rounded to 2 decimal places.

multiply the standard error by 2.33 and you get 2.33 * 8.5206 = 19.852998.

your sample mean is 48.76

the lower limit of your 98% confidence interval is 48.76 - 19.85 = 28.91.
the upper limit of your 98% confidence interval is 48.76 + 19.85 = 68.61.

this means that the probability that the true population mean is within these limits is 98%.

see http://onlinestatbook.com/2/estimation/mean.html for further reference.