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(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.

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Answer:

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

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