Chapter 11, Problem 6CRE

a.

To determine

To show: The provided data in a two-way table and conduct an appropriate chi-square test. Also, summarize the obtained results.

Answer to Problem 6CRE

There is insufficient evidence to support the claim of association.

Explanation of Solution

Given:

Number of consumers = 120

Number of students that said Yes = 22

Number of students that said No = 68

Formula used:

The formula to compute the chi-square test statistic is:

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

Using the provided information, the table is constructed as:

	Yes	No	Total
Student	22	39	61
Non-student	30	29	59
Total	52	68	120

Using this table, the expected frequencies are:

	Yes	No	Total
Student	26.4333	34.5667	61
Non-student	25.5667	33.4333	59
Total	52	68	120

The hypotheses are:

  $\begin{array}{l}{H}_0:\text{No association exists between the group}\\ H_a:\text{An association exists between the group}\end{array}$

The chi-square test statistic could be calculated as:

  $\begin{array}{c}{\chi }^2=\frac{{\left(22-26.433\right)}^2}{26.433}+\frac{{\left(36-34.5667\right)}^2}{34.5667}+\mathrm{......}+\frac{{\left(29-34.5667\right)}^2}{33.4333}\\ =2.6687\end{array}$

The degree of freedom is:

  $\begin{array}{c}\text{Degree of freedom =}\left(\text{Number of rows-1}\right)\left(\text{Number of columns-1}\right)\\ =\left(2-1\right)\left(2-1\right)\\ =1\end{array}$

The p-value using chi-square table at 6 degree of freedom is 0.102

The p-value is above significance level. The null hypothesis does not get rejected. There is enough evidence to be in favor of claim at 5% significance level.

b.

To determine

To analyze: The results of above part using the z-method. Also, check if the chi-square test statistic is the square of z-statistic.

Answer to Problem 6CRE

The chi-square test statistic is the square of z-statistic.

Explanation of Solution

Calculation:

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportion of students and non-students are:

  $\begin{array}{l}{\widehat{p}}_1=\frac{22}{61}=0.3607\\ {\widehat{p}}_2=\frac{30}{59}=0.5085\end{array}$

The output obtained using the Ti-83 plus calculator is:

  

From the above output, the p-value is 0.1023.

The p-value is above significance level. The null hypothesis could not be rejected. There is not enough evidence to conclude that there is a difference in two population proportions at 5% significance level.

Now, the verification for the chi-square test statistic could be done as:

Here, z-test statistic is -1.634.

Take square of z- statistic as:

  $\begin{array}{c}\left(z-statistic\right)={\left(1.634\right)}^2\\ =2.669\end{array}$

Here, obtained value is same as the chi-square test statistic. Thus, it could be said that the chi-square test statistic is the square of z-statistic.