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According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? State the random variable, population parameter, and hypotheses.

Respuesta :

Testing the hypothesis, it is found that since the p-value of the test is of 0.3015, which is greater than the significance level of 0.05, the data does not provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%.

At the null hypothesis, we test if the proportion is of 23%, that is:

[tex]H_0: p = 0.23[/tex]

At the alternative hypothesis, we test if this proportion is lower than 23%, that is:

[tex]H_1: p < 0.23[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

  • [tex]\overline{p}[/tex] is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

In this problem, the parameters are: [tex]p = 0.23, n = 1432, \overline{p} = \frac{321}{1432} = 0.2242[/tex]

Then, the value of the test statistic is:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.2242 - 0.23}{\sqrt{\frac{0.23(0.77)}{1432}}}[/tex]

[tex]z = -0.52[/tex]

The p-value of the test is the probability of finding a sample proportion of 0.2242 or below, which is the p-value of z = -0.52.

  • Looking at the z-table, z = -0.52 has a p-value of 0.3015.

Since the p-value of the test is of 0.3015, which is greater than the significance level of 0.05, the data does not provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%.

A similar problem is given at https://brainly.com/question/14639462