Get a similar question You can retry this question below According to a report on consumer fraud and identity theft, \( 20 \% \) of all complaints are for identity theft. This year, Delaware had 264 complaints of identity theft out of 1452 consumer complaints. Does this data provide enough evidence to show that Delaware has a lower proportion of identity theft than \( 20 \% \) ? Test at the \( 10 \% \) level. State the hypotheses. \( H_{0}: p=0 \) \( H_{a}: p \) Calculate the sample proportion. Round to four decimal places. \( \hat{p}=0.1818 \) Calculate the test statistic. Round to three decimal places. \( z=-1.734 \) Find the p-value. Round to four decimal places. \( p \)-value \( = \)

To determine if the data provides enough evidence to show that Delaware has a lower proportion of identity theft than 20%, we need to test the hypothesis at the 10% level. Given: - Total complaints in Delaware: 1452 - Complaints of identity theft in Delaware: 264 - National proportion of identity theft complaints: 20% (0.20) Hypotheses: \( H_{0}: p = 0.20 \) (The proportion of identity theft complaints in Delaware is equal to the national proportion) \( H_{a}: p < 0.20 \) (The proportion of identity theft complaints in Delaware is less than the national proportion) Sample proportion: \( \hat{p} = \frac{264}{1452} \approx 0.1818 \) Test statistic: \( z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \) where \( n = 1452 \) and \( p = 0.20 \) Substitute the values into the formula to calculate the test statistic: \( z = \frac{0.1818 - 0.20}{\sqrt{\frac{0.20(1-0.20)}{1452}}} \) Calculate the test statistic: \( z \approx -1.734 \) Now, we need to find the p-value associated with the test statistic. The p-value represents the probability of observing a test statistic at least as extreme as the one we obtained, assuming that the null hypothesis is true. The p-value is the probability of observing a z-score less than -1.734, which can be found using a standard normal distribution table or calculator. After calculating the p-value, we can compare it to the significance level (10%) to determine if the data provides enough evidence to reject the null hypothesis. Calculate the value by following steps: - step0: Calculate: \(0.1818-0.2\) - step1: Subtract the numbers: \(-0.0182\) Calculate or simplify the expression \( \sqrt{\frac{0.20(1-0.20)}{1452}} \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{\frac{0.2\left(1-0.2\right)}{1452}}\) - step1: Subtract the numbers: \(\sqrt{\frac{0.2\times 0.8}{1452}}\) - step2: Multiply the numbers: \(\sqrt{\frac{0.16}{1452}}\) - step3: Divide the terms: \(\sqrt{\frac{1}{9075}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{1}}{\sqrt{9075}}\) - step5: Simplify the expression: \(\frac{1}{55\sqrt{3}}\) - step6: Multiply by the Conjugate: \(\frac{\sqrt{3}}{55\sqrt{3}\times \sqrt{3}}\) - step7: Multiply the numbers: \(\frac{\sqrt{3}}{165}\) Calculate or simplify the expression \( \frac{0.1818 - 0.20}{\sqrt{\frac{0.20(1-0.20)}{1452}}} \). Calculate the value by following steps: - step0: Calculate: \(\frac{0.1818-0.2}{\sqrt{\frac{0.2\left(1-0.2\right)}{1452}}}\) - step1: Subtract the numbers: \(\frac{0.1818-0.2}{\sqrt{\frac{0.2\times 0.8}{1452}}}\) - step2: Multiply the numbers: \(\frac{0.1818-0.2}{\sqrt{\frac{0.16}{1452}}}\) - step3: Divide the terms: \(\frac{0.1818-0.2}{\sqrt{\frac{1}{9075}}}\) - step4: Simplify the root: \(\frac{0.1818-0.2}{\frac{\sqrt{3}}{165}}\) - step5: Subtract the numbers: \(\frac{-0.0182}{\frac{\sqrt{3}}{165}}\) - step6: Convert the expressions: \(\frac{-\frac{91}{5000}}{\frac{\sqrt{3}}{165}}\) - step7: Multiply by the reciprocal: \(-\frac{91}{5000}\times \frac{165}{\sqrt{3}}\) - step8: Reduce the numbers: \(-\frac{91}{1000}\times \frac{33}{\sqrt{3}}\) - step9: Multiply the fractions: \(-\frac{91\times 33}{1000\sqrt{3}}\) - step10: Multiply: \(-\frac{3003}{1000\sqrt{3}}\) - step11: Multiply by the Conjugate: \(\frac{-3003\sqrt{3}}{1000\sqrt{3}\times \sqrt{3}}\) - step12: Multiply the numbers: \(\frac{-3003\sqrt{3}}{3000}\) - step13: Reduce the fraction: \(\frac{-1001\sqrt{3}}{1000}\) - step14: Calculate: \(-\frac{1001\sqrt{3}}{1000}\) The test statistic is approximately -1.734. Now, we need to find the p-value associated with the test statistic. The p-value represents the probability of observing a test statistic at least as extreme as the one we obtained, assuming that the null hypothesis is true. The p-value is the probability of observing a z-score less than -1.734, which can be found using a standard normal distribution table or calculator. After calculating the p-value, we can compare it to the significance level (10%) to determine if the data provides enough evidence to reject the null hypothesis.