Answer to Question #292290 in Statistics and Probability for Yaku

Question

22. For a chi-square distribution,find x²_\alpha

such that

a) P(X² > x²_\alpha) = 0:99 when df = 4;

b) P(X² > x²_\alpha) = 0:025 when df = 19;

c) P(37.652 < X² < x²_\alpha) = 0:045 when df = 25.

23. Find

a) t_0.025 when df = 14

b) -t_0.10 when df = 10

c) t_0.995 when df = 7.

24. Find

a) P(T < 2:365) when df = 7;

b) P(-1.356 < T < 2.179) when df = 12;

c) P(T > 1.318) when df = 24;

d) P(T > -2.567) when df = 17.

25. For an F-distribution, find the value of f such that

a) P(F > f) = 0:05 when df₁ = 4 and df₂ = 9;

b) P(F > f) = 0:05 when df₁ = 9 and df₂ = 4;

c) P(F < f) = 0:95 when df₁ = 5 and df₂ = 8;

d) P(f₁ < F < f₂) = 0:90 when df₁ = 3 and df₂ = 9.

Answer 1

22

$a)$

$P(\chi^2 > \chi^2_\alpha) = 0.99$ when $df=4$

Here we look at $\alpha = 0.99$ and $df= 4$ to find from $\chi^2$ tables.,

$P(X^2 > 0.297) = 0.99$

$χ^ 2_{\alpha} = \chi^ 2_{ 0.99 }= 0.297$ for $df = 4$

$b)$

$P(X^2 \gt \chi^2_{\alpha}) = 0.025$ when df = 19

Here we look at $\alpha =0.025$ and $df=19$ to find,

$P(X^2 \gt 32.852) = 0.025$

So,

$\chi^2_{\alpha}=\chi^2_{0.025}=32.852$

$c)$

$P(37.652 \lt \chi^2 \lt \chi^2_{\alpha}) = 0.045$  when df = 25

Here we rewrite the inequality in a way that we can use the table. We Note that

$P(\chi^ 2_a \lt \chi^2 \lt \chi^2_b ) = P(\chi^ 2 \gt \chi^2_a ) − P(\chi^ 2 \gt \chi^2 _b ).$

So,

$P(37.652 \lt \chi^2 \lt \chi^2_{\alpha} ) = P(\chi^ 2 \gt 37.652) − P(\chi^ 2 \gt \chi^2_{\alpha} ) = 0.045.$

 First we look at the table for df= 25 and identify the $\chi^ 2_ {\alpha}$  value for 37.652 and find,

$\chi^ 2_{ 0.05} = 37.652$  for df = 25.

Thus,

$0.05 − P(\chi^ 2 \gt \chi^2_{\alpha}) = 0.045$

or

$P(\chi^ 2 \gt \chi^2_{\alpha}) = 0.005.$

Now we use the table for $\alpha= 0.005$ and $df = 25$ to find,

 $\chi^2_{ 0.005} = 46.928$ for $df = 25.$

23

$a)$

$t_{0.025 }$ when $df = 14$

From the t distribution table values, the t-value with $df=14$ leaving an area of 0.025 to the right is 2.145.

Therefore,

$t_{{0.025,14}}=2.145$

$b)$

$-t_{0.10}$ when $df = 10$

We apply the property, $-t_{\alpha}=t_{1-\alpha}$.

For $-t_{0.10,10}$  we can write, $-t_{0.10,10}=t_{1-0.10,10}=t_{0.90,10}$

From tables, we have that,

$t_{0.90,10}=-1.372184$ 

$c)$

$t_{0.995}$ when $df = 7$

The value required here is the t distribution table value with $df = 7$ that leaves an area of 0.995 to the right. From the tables, $t_{0.995,7}= -3.499483$

24

$a)$

$P(T \lt 2.365)$ when $df=7$

To use the table, we must transform the inequality to,

$P(T \lt 2.365) = 1 − P(T \gt 2.365).$

Using the table for $df = 7$ we see,

$t_{0.025} = 2.365,$ for $df= 7.$

Thus,

$P(T \lt 2.365) = 1 − 0.025 = 0.975.$

$b)$

$P(-1.356 \lt T \lt 2.179)$ when $df = 12$

Here we use symmetry for the fact that,

$P(T \lt −t_{\alpha}) = P(T \gt t_{\alpha}).$

Thus, we can write,

$P(−1.356 \lt T \lt 2.179) = 1−P(−1.356 \gt T)−P(T \gt 2.179) = 1−P(T \gt 1.356)−P(T \gt 2.179).$

Looking up these values in the table for $df = 12$ yields,

$t_{0.10} = 1.356$, and $t_{0.025} = 2.179$ for $df = 12$ . 

Therefore,

$P(−1.356 \lt T \lt 2.179) = 1 − 0.10 − 0.025 = 0.875.$

$c)$

$P(T \gt 1.318)$ when $df = 24.$

Using the table directly for $df = 24$ yields,

$t_{0.10} = 1.318$

Thus,

$P(T \gt 1.318) = 0.10.$

$d)$

$P(T \gt -2.567)$ when $df = 17$

Again, so we can use the table we write,

$P(T \gt −2.567) = 1 − P(T \lt −2.567).$

Then, symmetry gives,

$P(T \gt −2.567) = 1 − P(T \lt −2.567) = 1 − P(T \gt 2.567)$

The table gives,

$t_{0.01} = 2.567$ for $df = 17$

Therefore, $P(T\gt-2.567)=1-0.01=0.99$

25

$a)$

 $P(F \gt f) = 0.05$ when $df_1 = 4$ and $df2 = 9$ ;

Here, we look at the F distribution table value with numerator degrees of freedom equal to 4 and denominator degrees of freedom equal to 9 that leaves an area of 0.05 to the right given as,

$f_{0.05,4,9}=3.86$

So,

$P(F \gt 3.86) = 0.05$ when $df_1=4$ and $df_2=9$

$b)$

$P(F \gt f) = 0.05$ when $df_1 = 9$ and $df_2 = 4$ ;

Here, we look at the F distribution table value with numerator degrees of freedom equal to 9 and denominator degrees of freedom equal to 4 that leaves an area of 0.05 to the right given as,

$f_{0.05,9,4}=6.00$

Therefore,

$P(F\gt6.00)=0.05$ when $df_1=9$ and $df_2=4$

$c)$

$P(F \lt f) = 0.95$ when $df_1 = 5$ and $df_2 = 8$;

Here, $f$ is a value with numerator degrees of freedom equal to 5 and denominator degrees of freedom equal to 8 that leaves an area of 0.95 to the left of the F distribution. So, this value is same as the value with numerator degrees of freedom equal to 5 and denominator degrees of freedom equal to 8 that leaves an area of 0.05 to the right of the F distribution. So, $f_{0.05,5,8}=3.69$.

Therefore,

$P(F\lt 3.69)=0.95$ when $df_1=5$ and $df_2=8$

$d)$

$P(f_1 \lt F \lt f_2) = 0.90$ when $df_1 = 3$ and $df_2 = 9$

We rewrite this as, $P(f_1 \lt F \lt f_2)=P(F\gt f_1)-P(F\gt f_2)$ where, the value $f_1$ is the table value with $df_1=3$ and $df_2=9$ that leaves an area of 0.05 to the right or an area of 0.95 to the left, that is, $f_{0.95,3,9}$ and $f_2$ is the table value with $df_1=3$ and $df_2=9$ that leaves an area of 0.05 to the right, that is, $f_{0.05,3,9}$.

So,

$f_2=f_{0.05,3,9}=3.86$ and $f_1=f_{0.95,3,9}={1\over f_{0.05,9,3}}={1\over 8.81}=0.11$

Therefore,

$P(0.11 \lt F \lt 3.86) = 0.90$ when $df_1=3$ and $df_2=9.$