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 <u>test if the proportion is of 23%</u>, that is:
![H_0: p = 0.23](https://tex.z-dn.net/?f=H_0%3A%20p%20%3D%200.23)
At the alternative hypothesis, we <u>test if this proportion is lower than 23%</u>, that is:
![H_1: p < 0.23](https://tex.z-dn.net/?f=H_1%3A%20p%20%3C%200.23)
The test statistic is given by:
In which:
is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are: ![p = 0.23, n = 1432, \overline{p} = \frac{321}{1432} = 0.2242](https://tex.z-dn.net/?f=p%20%3D%200.23%2C%20n%20%3D%201432%2C%20%5Coverline%7Bp%7D%20%3D%20%5Cfrac%7B321%7D%7B1432%7D%20%3D%200.2242)
Then, the value of the <u>test statistic</u> is:
![z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B%5Coverline%7Bp%7D%20-%20p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D)
![z = \frac{0.2242 - 0.23}{\sqrt{\frac{0.23(0.77)}{1432}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.2242%20-%200.23%7D%7B%5Csqrt%7B%5Cfrac%7B0.23%280.77%29%7D%7B1432%7D%7D%7D)
![z = -0.52](https://tex.z-dn.net/?f=z%20%3D%20-0.52)
The p-value of the test is the probability of finding a sample proportion of 0.2242 or below, which is the <u>p-value of z = -0.52</u>.
- 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 brainly.com/question/14639462