Chapter 10.1, Problem 7CYU

a.

To determine

Find the expected frequencies for testing

$H_0:p_1=0.3,p_2=0.25,p_3=0.2,p_4=0.15,p_5=0.1$.

Answer to Problem 7CYU

The expected frequencies are,

Categories	Expected frequencies
1	21
2	17.5
3	14
4	10.5
5	7

Explanation of Solution

Calculation:

The observed frequencies of five categories are given.

Expected frequencies:

The expected frequencies are defined as $E_1=n{p}_1$, $E_2=n{p}_2$ and so on, for the probabilities of $p_1$, $p_2$,… are specified by null hypothesis $H_0$ with the total number of trial n.

Let $p_1$, $p_2$,…, $p_5$ denote the probability of each of the five categories.

The total number of trial is obtained as,

$\begin{array}{c}n=25+14+23+6+2\\ =70\end{array}$.

Now, the probabilities specified by $H_0$ are $p_1=0.3,p_2=0.25,p_3=0.2,p_4=0.15,p_5=0.1$ and the total number of trails is $n=70$.

Thus, the expected frequencies are,

Categories	Expected frequencies$\left(E_i=n{p}_i,\text{ for all }i=1,2,\mathrm{...},5\right)$
1	$\left(70\right)\left(0.3\right)=21$
2	$\left(70\right)\left(0.25\right)=17.5$
3	$\left(70\right)\left(0.2\right)=14$
4	$\left(70\right)\left(0.15\right)=10.5$
5	$\left(70\right)\left(0.1\right)=7$

b.

To determine

Explain whether the chi-square test is appropriate or not.

Answer to Problem 7CYU

The chi-square test is appropriate.

Explanation of Solution

It is given that one of the observed frequencies is less than 5.

It is known that when the null hypothesis $H_0$ is true the chi-square statistic follows approximately chi-square distribution provided that all the expected frequencies are 5 or more.

In part (a), it is found that all the expected frequencies corresponding to 5 categories are more than 5.

Hence, the chi-square test is appropriate.

c.

To determine

Find the value of ${\chi }^2$.

Answer to Problem 7CYU

The value of ${\chi }^2$ is 12.748.

Explanation of Solution

Calculation:

Chi-Square statistic:

The chi-square statistic is obtained as ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where $O_1,O_2,\mathrm{...},O_k$ be the observed frequencies and $E_1,E_2,\mathrm{...},E_k$ be the expected frequencies for k number of categories.

In the given question there are 5 categories.

The observed and expected frequencies are obtained as,

Categories	Observed frequencies	Expected frequencies
1	25	21
2	14	17.5
3	23	14
4	6	10.5
5	2	7

Now,

Categories	Observed frequencies (O)	Expected frequencies (E)	${\left(O-E\right)}^2$	$\frac{{\left(O-E\right)}^2}{E}$
1	25	21	16	0.7619
2	14	17.5	12.25	0.7
3	23	14	81	5.7857
4	6	10.5	20.25	1.9285
5	2	7	25	3.57143
Total	70	70	154.5	12.748

Thus, the value of ${\chi }^2$ is 12.748.

d.

To determine

Find the degrees of freedom.

Answer to Problem 7CYU

The degrees of freedom is 4.

Explanation of Solution

It is known that under the null hypothesis $H_0$ the test statistic ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ follows chi-square distribution with $k-1$ degrees of freedom for k number of categories, provided that all the expected frequencies are greater than or equal to 5.

Here, there are 5 categories. Thus, $k=5$.

Hence, the degrees of freedom is $5-1=4$.

Thus, the degrees of freedom is 4.

e.

To determine

Find the critical value at $\alpha =0.05$.

Answer to Problem 7CYU

The critical value at $\alpha =0.05$ is 9.488.

Explanation of Solution

Calculation:

From parts (c) and (d), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4.

Critical value:

In a test of hypotheses, the critical value is the point by which one can reject or accept the null hypothesis.

Critical value:

Software procedure:

Step-by-step software procedure to obtain critical value using MINITAB software is as follows:

• Select Graph>Probability distribution plot > view probability
• Select Chi -Square under distribution.
• In Degrees of freedom, enter 4.
• Choose Probability Value and Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.05 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

Hence, the critical value at $\alpha =0.05$ is 9.488.

f.

To determine

Explain whether the null hypothesis $H_0$ will be rejected at 0.05 level.

Answer to Problem 7CYU

The null hypothesis $H_0$ will be rejected at 0.05 level.

Explanation of Solution

Interpretation:

The hypotheses are:

Null Hypothesis:

$H_0:p_1=0.3,p_2=0.25,p_3=0.2,p_4=0.15,p_5=0.1$.

Alternate Hypothesis:

$H_1:\text{Not all }{p}_i\text{'}\text{s}\text{are same as }{H}_0\text{for }i=1,2,\mathrm{...},5$

From parts (c) and (e), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4 and the critical value is ${\chi }_{0.05,4}^2=9.488$.

  Rejection rule:

If the ${\chi }^2$ value is greater than or equal to critical value, that is ${\chi }^2\ge {\chi }_{\alpha ;k-1}^2$, then reject the null hypothesis. Otherwise, do not reject $H_0$.

Conclusion:

Here, the ${\chi }^2$ value is greater than to critical value.

That is, ${\chi }^2\left(=12.748\right)\ge {\chi }_{0.05,4}^2\left(=9.488\right)$

Thus, the decision is “reject the null hypothesis”.

Therefore, the null hypothesis $H_0$ will be rejected at 0.05 level.

g.

To determine

Find the critical value at $\alpha =0.01$.

Answer to Problem 7CYU

The critical value at $\alpha =0.01$ is 13.28.

Explanation of Solution

Calculation:

From part (c) and (d), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4.

Critical value:

Software procedure:

Step-by-step software procedure to obtain critical value using MINITAB software is as follows:

• Select Graph>Probability distribution plot > view probability
• Select Chi -Square under distribution.
• In Degrees of freedom, enter 4.
• Choose Probability Value and Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.01 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

Hence, the critical value at $\alpha =0.01$ is 13.28.

h.

To determine

Explain whether the null hypothesis $H_0$ will be rejected at 0.01 level.

Answer to Problem 7CYU

The null hypothesis $H_0$ is not rejected at 0.01level.

Explanation of Solution

Interpretation:

The hypotheses are:

Null Hypothesis:

$H_0:p_1=0.3,p_2=0.25,p_3=0.2,p_4=0.15,p_5=0.1$.

Alternate Hypothesis:

$H_1:\text{Not all }{p}_i\text{'}\text{s}\text{are same as }{H}_0\text{for }i=1,2,\mathrm{...},5$

From part (c) and (g), it is found that the value of chi-square test statistic is 12.748 with degrees of freedom 4 and the critical value is ${\chi }_{0.01,4}^2=13.28$.

  Rejection rule:

If the ${\chi }^2$ value is greater than or equal to critical value, that is ${\chi }^2\ge {\chi }_{\alpha ;k-1}^2$, then reject the null hypothesis. Otherwise, do not reject $H_0$.

Conclusion:

Here, the ${\chi }^2$ value is less than the critical value.

That is, ${\chi }^2\left(=13.28\right)<{\chi }_{0.01,4}^2\left(=13.28\right)$

Thus, the decision is “fail to reject the null hypothesis”.

Therefore, the null hypothesis $H_0$ is not rejected at 0.01level.