Q.6 - The Practice of Statistics for AP

Question

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.

Step-by-Step Solution

Answer

(a) The value of p is above significant level

b) As per the results we can conclude that the chi-square test statistic is the square of z-statistic.

Part (a) step 1 : Given information

No of consumers=$120$

No of students agreed =width="19" style="max-width: none; vertical-align: -4px;" $22$

No of students  non agreed =$68$

Part (a) step2 : calculation

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

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

From given information, table as follows:

	agreed	non agreed	total
student	$22$	$39$	$61$
non student	$30$	$29$	$59$
total	$52$	width="19" style="max-width: none; vertical-align: -4px;" $68$	$120$

 

From this expected frequencies:

	agreed	non  agreed	total
student	$26.4333$	$34.5667$	$61$
non student	$25.5667$	$33.4333$	$59$
total	$52$	$68$	$120$

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:

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

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.

Part (b) step 1: Given information

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

Part (b) step2 : calculation

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:

$P1=\frac{22}{61}=0.3607$

$P2=\frac{30}{59}=0.5085$

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 the z-statistic.