Chapter 14.4, Problem 14.18E

i.

To determine

No of degrees of freedom associated with $x^2$ statistic.

Answer to Problem 14.18E

No of degrees of freedom associated with $x^2$ statistic is $2$ .

Explanation of Solution

Given:

Columns
Rows	$1$	$2$	$3$	Total
$1$	$37$	$34$	$93$	$164$
$2$	$66$	$57$	$113$	$236$
Total	$103$	$91$	$206$	$400$

Formula Used:

Degrees of freedom for contingency table:

  $df=\left(r-1\right)\left(c-1\right)$

Calculation:

Consider null and alternative hypothesis.

Null hypothesis, $H_0$ : Response falls in any one row is independent of the column if falls in.

Alternative hypothesis, $H_1$: Response falls in any one row is dependent of the column if falls in.

To find degrees of freedom for contingency table:

  $df=\left(r-1\right)\left(c-1\right)$

Where,

  $r=$ no of rows.

  $c=$ no of columns.

So, the contingency table contains two rows and three columns.

Degree of freedom,

  $\begin{array}{l}df=\left(r-1\right)\left(c-1\right)\\ =\left(2-1\right)\left(3-1\right)\\ =1\times 2\\ =2\end{array}$

Thus, $2$ degrees of freedom are associated with the Chi-square test statistic.

Conclusion:

Thus, no. of degrees of freedom associated with $x^2$ statistic is $2$ .

ii.

To determine

To find: the value of test statistic.

Answer to Problem 14.18E

The value of test statistic is $3.059$ .

Explanation of Solution

Given:

Columns
Rows	$1$	$2$	$3$	Total
$1$	$37$	$34$	$93$	$164$
$2$	$66$	$57$	$113$	$236$
Total	$103$	$91$	$206$	$400$

Formula Used:

Test statistic:

  $Ef=\frac{R_t\times C_t}{T}$

Where, $Ef=$ expected frequency.

  $R_t=$ Row total.

  $C_t=$ Column total.

  $T=$ Grand total.

Calculation:

Observed$O_i$	Expected $E_i$	$\left(O_i-E_i\right)$	${\left(O_i-E_i\right)}^2$	$x^2=\frac{{\left(O_i-E_i\right)}^2}{E_i}$
$37$	$\frac{164\times 103}{400}=42.23$	$-5.23$	$27.353$	$0.648$
$34$	$\frac{164\times 91}{400}=37.31$	$-3.31$	$10.956$	$0.294$
$93$	$\frac{164\times 206}{400}=84.46$	$8.54$	$72.932$	$0.864$
$66$	$\frac{236\times 103}{400}=60.77$	$5.23$	$27.353$	$0.450$
$57$	$\frac{236\times 91}{400}=53.69$	$3.31$	$10.956$	$0.204$
$113$	$\frac{236\times 206}{400}=121.54$	$-8.54$	$72.932$	$0.600$
$400$	$400$

From the above table, the test statistic which is observed is $3.059$ .

Conclusion:

Thus, $3.059$ is the observed test statistic.

iii.

To determine

To find:

Rejection region for $\alpha =0.01$ .

Answer to Problem 14.18E

the test statistic $\left(x_{0.01}^2\right)$ is $9.21034$ with level of significance $\alpha =0.01$ and degrees of freedom $df=2$ .

Explanation of Solution

Given:

  $\alpha =0.01=$ level of significance.

Columns
Rows	$1$	$2$	$3$	Total
$1$	$37$	$34$	$93$	$164$
$2$	$66$	$57$	$113$	$236$
Total	$103$	$91$	$206$	$400$

Calculation:

The given level of significance $\left(\alpha \right)$ is $0.01$

The rejection region,

  $RR=\left\{x^2>x_{0.01}^2|rejectthe{H}_0\right\}$

The critical value with $\alpha =0.01$ and degrees of freedom $df=2$ is obtained from the chi-square table as $x_{0.01}^2=9.21034$ .

Conclusion:

Thus, the rejection region $\left(x_{0.01}^2\right)$ is $9.21034$ with level of significance $\alpha =0.01$ and degrees of freedom $df=2$ .

iv.

To determine

To identify: test and its conclusion.

Answer to Problem 14.18E

The probability that a response falls in any row is independent of the columns if falls in.

Explanation of Solution

Given:

Columns
Rows	$1$	$2$	$3$	Total
$1$	$37$	$34$	$93$	$164$
$2$	$66$	$57$	$113$	$236$
Total	$103$	$91$	$206$	$400$

Calculation:

For getting test initially compare the calculated value with the critical value.

By using the rejection region, the calculated value of test statistic less than the level of significance than fails to reject the null hypothesis.

Conclusion:

Thus, the probability that a response falls in any row is independent of the columns if falls in.

v.

To determine

To identify: The P-value for the test.

Answer to Problem 14.18E

The P-value for the test is $0.217$ .

Explanation of Solution

Given:

Columns
Rows	$1$	$2$	$3$	Total
$1$	$37$	$34$	$93$	$164$
$2$	$66$	$57$	$113$	$236$
Total	$103$	$91$	$206$	$400$

Calculation:

Critical value is calculated as:

P-value $=0.217$ (by using Excel’s Chidist (probability, deg_freedom)).

So, here $x=3.059$ and degrees of freedom $df=2$ .

Conclusion:

Thus, the required P-value is $0.217$ .