Question

The hourly rate of substitute teachers for 12 local school districts is given below. Assuming that the data are normally distributed, use a TI-83, or TI-84 calculator to find the 90% confidence interval for the mean hourly rate of substitute teachers in the region.20 13 21 18 19 2219 15 12 12 18 21

Answer 1

`17.5-1.796(3.61)/(√(12))=15.63`

`17.5+1.796(3.61)/(√(12))=19.37`

Explanation:

Data given

20 13 21 18 19 22 19 15 12 12 18 21

We can calculate the sample mean and deviation with the following formulas:

`\bar X =(\sum_(i=1)^n X_i)/(n)`

`s= \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And we got:

`\bar X = 17.5` represent the sample mean

`\mu` population mean (variable of interest)

s=3.61 represent the sample standard deviation

n=12 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=12-1=11`

Since the Confidence is 0.90 or 90%, the significance is
`\alpha=0.1` and
`\alpha/2 =0.05`, the critical value would be given by
`t_(\alpha/2)=`

Now we have everything in order to replace into formula (1):

`17.5-1.796(3.61)/(√(12))=15.63`

`17.5+1.796(3.61)/(√(12))=19.37`