Question

The grade point averages for 10 randomly selected junior college students are 2.0, 3.2, 1.8, 2.9, 0.9, 4.0, 3.3, 2.9, 3.6, 0.8. Assume the grade point averages are normally distributed. What is the 98% confidence interval for the true mean?

Answer 1

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution to find that with `98%` confidence that the mean lies in the range

`1.55, 3.53`

Explanation:

One way to calculate the confidence interval of a sample mean is to use the student's t-distribution where the limits of the interval are given by:

`m+- t_(1-alpha//2,N-1)s/sqrt(N)`

where

`m` is the sample mean
`alpha` is the residual probability;in our case `alpha = 1-0.98 = 0.02`
`s` is the sample standard deviation
`N` is the number of samples (`N-1` is the number of degrees of freedom)

First we need the sample mean and sample standard deviation:

`m=1/N sum_(i=1)^N x_i=2.54`

and the sample standard deviation:

`s = sqrt(1/(N-1) sum_(i=1)^N(x_i-m)^2) =1.11`

Plugging these values into our first equation we can calculate the limits of the confidence interval for our mean:

`2.54+- t_(0.99,9)1.11/sqrt(10)`

Where we can look up the value of `t_(0.99,9)` in a table or use a function in a software package, like the T.INV function in Microsoft Excel to find the value `2.82`

`2.54+-2.82*1.11/sqrt(10)`

therefore we can say with `98%` confidence that the mean lies in the range

`1.55, 3.53`