Question

A study of identity theft looked at how well consumers protect themselves from this increasingly prevalent crime. The behaviors of 61 college students were compared with the behaviors of 55 nonstudents. 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 26 of the nonstudents agreed.

(a) Display the data in a two-way table.

	Students		Nonstudents		Total
Agreed
Disagreed
Total					116

Perform the chi-square test. (Round your χ² to three decimal places and round your P-value to four decimal places.)

χ2	=
df	=
P-value	=

Summarize the results.

We cannot conclude at the 5% level that students and nonstudents differ in the response to this question.We can conclude at the 5% level that students and nonstudents differ in the response to this question.    

(b) Reanalyze the data using the methods for comparing two proportions that we studied in the previous chapter. Compare the results and verify that the chi-square statistic is the square of the z statistic. (Test students who agreed minus nonstudents who agreed. Round your z to two decimal places and round your P-value to four decimal places.)

z	=
P-value	=

(c) The students in this study were junior and senior college students from two sections of a course in Internet marketing at a large northeastern university. The nonstudents were a group of individuals who were recruited to attend commercial focus groups on the West Coast conducted by a lifestyle marketing organization. Discuss how the method of selecting the subjects in this study relates to the conclusions that can be drawn from it.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: p₁ = p₂
H_a: At least one of the null hypothesis statements is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1)
D.F = 1
E_r,c = (n_r * n_c) / n

	Students	Expected	[(Or,c - Er,c)2/ Er,c]	Non-students	Expected	[(Or,c - Er,c)2/ Er,c]	Total
Agreed	22	25.2414	0.416242698	26	22.75862069	0.461650993	48
Disagreed	39	35.7586	0.293818375	29	32.24137931	0.325871289	68
Total	61	61	0.710061074	55	55	0.787522282	116

Χ² = 1.498

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 1.50.

We use the Chi-Square Distribution Calculator to find P(Χ² > 1.498) = 0.221

Interpret results. Since the P-value (0.221) is greater than the significance level (0.05), we have to reject the null hypothesis.

We cannot conclude at the 5% level that students and non students differ in the response to this question.    

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.41379
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.09158
z = (p₁ - p₂) / SE

z = - 1.22

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.22 or greater than 1.22.

Thus, the P-value = 0.222

Interpret results. Since the P-value (0.222) is greater than the significance level (0.05), hence we failed to reject the null hypothesis.

We cannot conclude at the 5% level that students and non students differ in the response to this question.    

Hence approximately z² is equal to chi square statistics.