Question

(4 points) Identify the critical t. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t value t∗ for the given sample size and confidence level. Round critical t values to 4 decimal places.

Answer 1

a)
`df =n-1= 5-1= 4`

For this case the significance level is given by
`\alpha=1-0.9=0.1` and the
`\alpha/2 =0.05` and for this case we can ue the following code:

"=-T.INV(0.05,4)"

And for this case the critical value would be
`t_(cric)= 2.1319`

b)
`df =n-1= 13-1= 12`

For this case the significance level is given by
`\alpha=1-0.95=0.05` and the
`\alpha/2 =0.025` and for this case we can ue the following code:

"=-T.INV(0.025,12)"

And for this case the critical value would be
`t_(cric)=2.1788`

c)
`df =n-1= 22-1= 21`

For this case the significance level is given by
`\alpha=1-0.98=0.02` and the
`\alpha/2 =0.01` and for this case we can ue the following code:

"=-T.INV(0.01,21)"

And for this case the critical value would be
`t_(cric)=2.5176`

d)
`df =n-1= 15-1= 14`

For this case the significance level is given by
`\alpha=1-0.99=0.01` and the
`\alpha/2 =0.005` and for this case we can ue the following code:

"=-T.INV(0.005,14)"

And for this case the critical value would be
`t_(cric)=2.9768`

Explanation:

Part a

n = 5 , Conf. =0.90

For this case the degrees of freedom are given by:

`df =n-1= 5-1= 4`

For this case the significance level is given by
`\alpha=1-0.9=0.1` and the
`\alpha/2 =0.05` and for this case we can ue the following code:

"=-T.INV(0.05,4)"

And for this case the critical value would be
`t_(cric)= 2.1319`

Part b

n = 13 , Conf. =0.95

For this case the degrees of freedom are given by:

`df =n-1= 13-1= 12`

For this case the significance level is given by
`\alpha=1-0.95=0.05` and the
`\alpha/2 =0.025` and for this case we can ue the following code:

"=-T.INV(0.025,12)"

And for this case the critical value would be
`t_(cric)=2.1788`

Part c

n = 22 , Conf. =0.98

For this case the degrees of freedom are given by:

`df =n-1= 22-1= 21`

For this case the significance level is given by
`\alpha=1-0.98=0.02` and the
`\alpha/2 =0.01` and for this case we can ue the following code:

"=-T.INV(0.01,21)"

And for this case the critical value would be
`t_(cric)=2.5176`

Part d

n = 15 , Conf. =0.99

For this case the degrees of freedom are given by:

`df =n-1= 15-1= 14`

For this case the significance level is given by
`\alpha=1-0.99=0.01` and the
`\alpha/2 =0.005` and for this case we can ue the following code:

"=-T.INV(0.005,14)"

And for this case the critical value would be
`t_(cric)=2.9768`