SOLUTION: please help me solve this question. a random sample of 40 shoppers showed that they spend an average of 23.45$ per visit at the saturday mornings bookstore. the standard deviation

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Question 933566: please help me solve this question.
a random sample of 40 shoppers showed that they spend an average of 23.45$ per visit at the saturday mornings bookstore. the standard deviation of the sample is 2.80$.find the 98% confidence interval of the true mean? find the 95% confidence interval of the true variance?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
sample mean is 23.45
sample standard deviation is 2.80.
sample size = 40

critical alpha for two tailed 98% confidence interval = (1-.98)/2 = .01
z factor for a tail of .01 is equal to z(1-.01) = z(.99) = 2.3263 rounded to 4 decimal places.

critical z-factor is equal to 2.3263 rounded to 4 decimal places.

standard error = standard deviation / square root of sample size = 2.80 / sqrt(40) = .4427 rounded to 4 decimal places.

margin of error = standard error * critical z factor = .4427 * 2.3263 = 1.0299 rounded to 4 decimal places.

confidence interval = mean plus or minus margin of error.

lower limit of confidence interval of the mean = 23.45 - 1.0299 = 22.4201.

upper limit of confidence interval of the mean = 23.45 + 1.0299 = 24.4799.

there is a 98% probability that the true mean will lie between 22.4201 and 24.4799.

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estimating the true variation requires a different formula as best i can determine.

the best reference i could find that i could semi-understand can be found at the following link:

http://www.milefoot.com/math/stat/ci-variances.htm>

the information i referenced from this link is found at the bottom of the link under example.

the sample standard deviation of the sample is equal to 2.8.

the sample size is 40.

the degrees of freedom = 40 - 1 = 39.

95% confidence interval give you a .025 tail on each end of the distribution curve.

you will be looking for the chi-square value for .025 and .975 with 39 degrees of freedom.

the chi-square calculator i used can be found at the following link:

http://stattrek.com/online-calculator/chi-square.aspx

chi-square value for .025 with 39 degrees of freedom = 58.1
chi-square value for .975 with 39 degrees of freedom = 23.7

the equation to be evaluated using these chi-square values is:

estimated population variance = ((n-1) * sample variance) / chi-square value

sample variance is equal to sample standard deviation squared is equal to 2.80 squared is equal to 7.84.

formula is evaluated at the given chi-square values to get:

population variance 1 = (39 * 7.84 / 58.1 = 5.2627

population variance 2 = (39 * 7.84) / 23.7 = 12.9013

the 95% confidence interval of the true variance has a lower limit of 5.2627 and an upper limit of 12.9013.

take the square root of those numbers to get the confidence interval of the true standard deviation.

i got population standard deviation estimated to be between 2.29 and 3.59.

you didn't want the population standard deviation, though.
you wanted the population variance.

that is estimated to be between 5.2627 and 12.9013.