256 middle managers were randomly selected and their annual income noted. The sample mean is $35,420.00 and the sample standard deviation is $2050.00.  a) What is the estimated mean income of all middle managers? b) What is the 89% confidence interval (rounded to the nearest $10)? c)Interpret your answer for (b)?

Expert Answers

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a) The sample mean of the data is `barx =` $35,420, so our estimated mean income for any middle manager in the population is exactly that, $35,420.

b) Assuming that the income of middle managers is Normally distributed, then the sampling distribution of the sample mean variable `barX` is such that

`bar X` ~ Normal(`mu`, `sigma^2/n` )

` `where `mu` is the mean income of a middle manager and `sigma^2` is the population variance. We are interested in a conifdence interval for the mean `mu`.

We estimate the mean of `bar X` (`mu`) by the sample mean `bar x` and the standard deviation of `bar X` (`sigma/sqrt(n)` ) by the sample standard deviation `hat (sigma)` divided by the square-root of the sample size `n`.

From our data we have that `hat(sigma) = $2050` so `hat(sigma)/sqrt(n) = 2050/sqrt(256) = 2050/16 = $128.125`

A `(1-alpha)` % confidence interval for the mean ` `income `mu` is given by the formula

`bar x +- Phi^(-1)(1-alpha/2)hat(sigma)/sqrt(n)`

where `Phi^(-1)` is the inverse cdf function for the standard Normal distribution.

Here we have `1-alpha = 0.89 implies alpha = 0.11 implies alpha/2 = 0.055`

Using a standard Normal look-up table `Phi^(-1)(1-alpha/2) = Phi^(-1)(0.945) = 1.598`

So a 89% confidence interval for `mu` is given by

`barx -+ 1.598 hat(sigma)/sqrt(n) = 35420 -+ 204.74 = [$35,220,$35,620]`

to the nearest $10

c) The 89% confidence interval for `mu` [$35,220,$35,620] contains the true mean income of middle managers `mu` 89% of the time (over repeated draws of samples from the population)

a) $35,420 b) [$35,220,$35,620] c) interval contains `mu` 89% of the time

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