Q 8.18RP. - Elementary Statistics

Question

For a $t-$curve with $\mathrm{df}=18$, obtain the $t-$value and illustrate your results graphically.

a. The $t-$value having area $0.025$ to its right

b. $t_{0.05}$

c. The $t-$value having area $0.10$ to its left

d. The two $t-$values that divide the area under the curve into a middle $0.99$ area and two outside $0.005$ areas

Step-by-Step Solution

Answer

Part (a) for $df=18, t_{0.025}=2.101$

Part (b) for $df=18,t_{0.05}=1.734$

Part (c) $t-$value having an area $0.10$ to its left $=-t_{0.10}$ $=-1.330$

Part (d) The required two $t-$values are $-2.878$ and $2.878$

Part (a) Step 1: Given information

Degrees of freedom of the $t-$ curve is $18$

Part (a) Step 2: Explanation

To obtain the $t-$ value having area $0.025$ to its right i.e., the value of $t_{0.025}$ for $df=18, t_{0.025}=2.101$

Part (b) Step 1: Explanation

To find the value of $t_{0.05}$ for $df=18, t_{0.05}=1.734$

Part (c) Step 1: Explanation

Given that the $t-$curve is symmetric about $0$, a $t-$value with an area of $0.10$ to the left is equal to the negative of a $t-$value with an area of $0.10$ to the right.

$\therefore$ for$d f=18$,

$t-$value having area $0.10$ to its left $=-t_{0.10}$ $=-1.330$

Part (d) Step 1: Explanation

It is necessary to determine the two $t-$ values that divide the curve's surface area into a $0.99$ central area and two $0.005$ outlying areas.

i.e., we have to obtain $-t_{0.005}$ and $t_{0.005}$ for $d f=18$

[$\therefore$-curve is symmetric about $0$]

For $df=18,t_{0.005}=2.878$

$\therefore$ The required two $t-$values are $-2.878$ and $2.878$

$0.005$