What is the answer to part B and C ? Please Help A study of identity theft looke

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What is the answer to part B and C ? Please Help

A study of identity theft looked at how well consumers protect themselves from this increasingly prevalent crime. The behavior of 64 college students were compared with the behaviors of 59 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 1st four digits of my social security number, or a series of consecutive numbers."For the students, 23 agreed with this statement while 30 of the nonstudents agreed. (a) Display the data in a two-way table. Perform the chi=square test. x^2 = 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 of the z statistic. z = P-value =

Explanation / Answer

Solution:-

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

Null hypothesis: PStudents - PNon-students = 0

Alternative hypothesis: PStudents - PNon-students 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

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 = 0.411

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.089

z = (p1 - p2) / SE

z = - 1.68

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 -1.68 or greater than 1.68

Thus, the P-value = 0.093

Interpret results. Since the P-value (0.093) is more than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that students and non sutudents differ in responses.

c) The two samples are independent of each other.

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