Q.6 A study of identity theft looked... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 11: Q.6 (page 733)

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 鈥淲hen asked to create a password, I have used either my mother鈥檚 maiden name, or my pet鈥檚 name, or my birth date,
or the last four digits of my social security number, or a series of consecutive numbers.鈥� 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.

Short Answer

Expert verified

(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 ofz-statistic.

Step by step solution

Part (a) step 1 : Given information

No of consumers= $120$

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

No of students non agreed = $68$

Part (a) step2 : calculation

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

$蠂^{2} = 鈭� \frac{(O 鈭� E)^{2}}{E}$

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;" localid="1650927989861" $68$	$120$

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:

localid="1650949427330" role="math" $\begin{matrix} 蠂^{2} = \frac{(22 鈭� 26.433)^{2}}{26.433} + \frac{(36 鈭� 34.5667)^{2}}{34.5667} + 鈥� . . . + \frac{(29 鈭� 34.5667)^{2}}{33.4333} \end{matrix} = 2.6687$

The degree of freedom is:

$Degree of freedom = (Number of rows-1) (Number of columns- 1) = (2 - 1) (2 - 1) = 1$

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:

$\begin{aligned} H_{0} : p 1 = p 2 \\ H_{a} : p 1 鈮� p 2 \end{aligned}$

The proportion of students and non-students are:

localid="1651488609427" role="math" $P 1 = \frac{22}{61} = 0.3607$

localid="1651488621820" $P 2 = \frac{30}{59} = 0.5085$

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.