A study of identity theft looked at how well consumers protect themselves from this increasingly prevalent crime. The behaviors of $61$randomly selected college students were compared with the behaviors of $59$randomly selected non students.$39$One of the questions was “When asked to create a password, I have used either my mother’s maiden name, or my pet’s name, or my birth date,

or the last four digits of my social security number, or a series of consecutive numbers.” For the students, $22$agreed with this statement while $30$of the nonstudents agreed.

a) Display the data in a two-way table and perform

the appropriate chi-square test. Summarize the results.

(b) Reanalyze the data using the methods for comparing two proportions that we studied in Chapter$10$. Compare the results and verify that the chi-square

statistic is the square of the z statistic.

No of consumers=$120$

No of students agreed =width="19">22

No of students non agreed =$68$

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

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

From given information, table as follows:

From this expected frequencies:

The hypotheses are:

H₀: No association exists between the group

H_a: An association exists between the group

The chi-square test statistic could be calculated as:

localid="1650949427330" role="math" $$\begin{gathered} \begin{array}{c}{\chi }^2=\frac{(22-26.433{)}^2}{26.433}+\frac{(36-34.5667{)}^2}{34.5667}+\dots ...+\frac{(29-34.5667{)}^2}{33.4333}\end{array} \\ =2.6687 \end{gathered}$$

The degree of freedom is:

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

The chi-square table p-value at $6$degrees of freedom is $0.102$.

The p-value is above the level of significance. At the $5\%$significance level, there is sufficient evidence to support the claim.

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

The null and alternative hypotheses are:

$\begin{array}{r}{H}_0:p1=p2\\ H_a:p1\ne p2\end{array}$

The proportion of students and non-students are:

localid="1651488609427" role="math" $P1=\frac{22}{61}=0.3607$

localid="1651488621820" $P2=\frac{30}{59}=0.5085$

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

The p-value is $0.1023$based on the above output.

The p-value is above the level of significance.

At the $5\%$significance level, there is insufficient evidence to establish that there is a difference in two population proportions. The chi-square test statistic might now be verified as follows:

The z-test statistic in this case is$-1.634$.

Take the square of the z-statistic as follows:

(z-statistic)=${(1.634)}^2$

$=2.669$

The resultant value is identical to the chi-square test statistic in this case. As a result, the chi-square test statistic is equal to the square of thez-statistic.