Question

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. Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%?

Answer 1

Answer:

There is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1432

p = 23% = 0.23

Alpha, α = 0.05

Number of theft complaints , x = 321

First, we design the null and the alternate hypothesis  

$H_{0}: p = 0.23\\H_A: p < 0.23$

This is a one-tailed test.  

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{321}{1432} =0.2241$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.2241-0.23}{\sqrt{\frac{0.23(1-0.23)}{1432}}} = -0.5305$

Now, we calculate the p-value from the table.

P-value = 0.298

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.