Step 1

(a)

Given,

The level of significance is $\alpha = 0.05$.

The degrees of freedom is 4.

The calculated value of the test statistic is, ${\chi ^2} = 12.25$.

The null and alternative hypotheses are follows,

\[\begin{array}{l} {H_0}:{p_1} = ... = {p_n}\\ {H_1}:{\rm{ Atleast\;one\;}}{p_i}{\rm{\;does\;not\;equal}}{\rm{.}} \end{array}\]

Now calculate the p-value by using significance level and degrees of freedom.

Therefore, the p-value is 0.0156.

It is clear that, the p-value is less than the value of significance level.

That is, $0.0156 < 0.05$.

Hence, reject the null hypothesis at significance level 0.05.

 
Step 2

(b)

Given,

The level of significance is $\alpha = 0.01$.

The degrees of freedom is 3.

The calculated value of the test statistic is, ${\chi ^2} = 8.54$.

The null and alternative hypotheses are follows,

\[\begin{array}{l} {H_0}:{p_1} = ... = {p_n}\\ {H_1}:{\rm{ Atleast\;one\; }}{p_i}{\rm{ \;does \;not \;equal}}{\rm{.}} \end{array}\]

Now calculate the p-value by using significance level and degrees of freedom.

Therefore, the p-value is 0.0361.

It is clear that, the p-value is greater than the value of significance level.

That is, $0.0361 > 0.01$.

Hence, fail to reject the null hypothesis at significance level 0.01.

 
Step 3

(c)

Given,

The level of significance is $\alpha = 0.10$.

The degrees of freedom is 2.

The calculated value of the test statistic is, ${\chi ^2} = 4.36$.

The null and alternative hypotheses are follows,

\[\begin{array}{l} {H_0}:{p_1} = ... = {p_n}\\ {H_1}:{\rm{ Atleast \;one\; }}{p_i}{\rm{ \;does \;not \;equal}}{\rm{.}} \end{array}\]

Now calculate the p-value by using significance level and degrees of freedom.

Therefore, the p-value is 0.1130.

It is clear that, the p-value is greater than the value of significance level.

That is, $0.1130 > 0.10$.

Hence, fail to reject the null hypothesis at significance level 0.10.

 
Step 4

(d)

Given,

The level of significance is $\alpha = 0.01$.

The number of observations is, $k = 6$.

The calculated value of the test statistic is, ${\chi ^2} = 10.20$.

The null and alternative hypotheses are follows,

\[\begin{array}{l} {H_0}:{p_1} = ... = {p_n}\\ {H_1}:{\rm{ Atleast \;one\; }}{p_i}{\rm{ \;does \;not \;equal\;}}{\rm{.}} \end{array}\]

Now, calculate the degrees of freedom is given below,

\[\begin{array}{c} df = k - 1\\ = 6 - 1\\ = 5 \end{array}\]

Also, calculate the p-value by using significance level and degrees of freedom.

Therefore, the p-value is 0.0698.

It is clear that, the p-value is greater than the value of significance level.

That is, $0.0698 > 0.01$.

Hence, fail to reject the null hypothesis at significance level 0.01.