**Analysis of Insurance Policy Distribution Changes**Over the past decade, an insurance company has maintained a mix of policies: 40% whole life policies, 25% universal life policies, 25% annual renewable-term (ART) policies, and 10% other types of policies. Changes in this mix over the long term could necessitate modifications in commission structures, reserves, and possibly investments. Below are the results from a random sample of 850 policies issued over the past few months:| Category | Number ||----------------|--------|| **Whole life** | 354 || **Universal life** | 215 || **ART** | 206 || **Other** | 75 |**Hypothesis Test**To determine if there is sufficient evidence, at the 5% level of significance, that the distribution of policy types has shifted from the historical percentages, you can conduct a hypothesis test. Follow these steps for the test:1. **State the hypotheses:** - Null hypothesis (H₀): The distribution of policy types is the same as the historical distribution. - Alternative hypothesis (H₁): The distribution of policy types is different from the historical distribution.2. **Calculate the expected frequencies:** - Based on the historical percentages: - Whole life: $0.40 \times 850 = 340$ - Universal life: $0.25 \times 850 = 212.5$ - ART: $0.25 \times 850 = 212.5$ - Other: $0.10 \times 850 = 85$3. **Calculate the chi-square statistic:** - $ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} $ where $O_i$ is the observed frequency, and $E_i$ is the expected frequency for each category.4. **Compare the chi-square statistic to the critical value from the chi-square distribution table with the appropriate degrees of freedom (df = number of categories - 1).**5. **Decision rule:** - If the calculated chi-square statistic is greater than the critical value, reject the null hypothesis.**Report your solution in the space below:** Be sure to address all points for a hypothesis test. ### Hypothesis Test RequirementsFor all questions involving a hypothesis test, be sure to include all the required steps:1. **State the null (H₀) and alternative (Hₐ or H₁) hypotheses and identify the claim.**2. **Using the sample data, compute the test statistic.**3. **Using the level of significance, calculate the critical value.**4. **Compute the p-value.**5. **Using the test statistic and critical value, make a decision: Reject H₀ or Fail to reject H₀.** - Using the p-value and level of significance, make a decision: Reject H₀ or Fail to reject H₀.6. **Interpret the results in the context of the original problem.**#### Decision-Making TableThe table below shows how to interpret the decisions based on the hypotheses:| | **Decision: Reject H₀** | **Decision: Fail to reject H₀** ||---------------------|--------------------------------------------------------------------------------|-------------------------------------------------------------------------------------|| **Claim: H₀** | "There is sufficient evidence to warrant rejection of the claim that ... (original claim)." | "There is not sufficient evidence to warrant rejection of the claim that ... (original claim)." || **Claim: Hₐ** | "The sample data support the claim that ... (original claim)." | "There is not sufficient sample evidence to support the claim that ... (original claim)." |This table helps clarify the interpretation of results depending on whether the null hypothesis (H₀) or the alternative hypothesis (Hₐ) is being tested. For example, if the null hypothesis is rejected, it suggests that there is enough evidence to support the alternative hypothesis. Conversely, if the null hypothesis fails to be rejected, it suggests that there isn't sufficient evidence to support the alternative hypothesis.By following these steps and interpretations, you can systematically perform hypothesis testing and clearly understand the results in the context of your specific problem.

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Answered: Over the past decade, an insurance…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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An insurance company offers four different deductible levels—none, low, medium, and high—for its homeowner's policyholders and three different levels—low, medium, and high—for its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner's deductible and low auto deductible is 0.07 (7% of all such individuals). Homeowner's Auto N L M H L 0.04 0.07 0.05 0.01 M 0.07 0.11 0.20 0.10 H 0.02 0.03 0.15 0.15 Suppose an individual having both types of policies is randomly selected. (a) What is the probability that the individual has a medium auto deductible and a high homeowner's deductible?(b) What is the probability that the individual has a low auto deductible? A low homeowner's deductible? auto deductible homeowner's deductible (c) What is the probability that the individual is in the same…
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The selling prices of mutual funds change daily. In order to study these changes, a sample of mutual funds was examined and the daily changes in price are listed below. (Round answers to 3 decimal places) 0.32, -0.17, 0.26, -0.03, -0.01, 0.18, 0.33, 0.28, 0.02, -0.29, -0.08, 0.12, 0.07, 0.03, 0.28 a) Using a calculator find Q1, Q3, median and IQR b) Determine the lower and upper fences. (Show work) c) Identify the outliers (if any) in this set
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Chapter 5 In preparation for your discussion post, complete the following: • Research data on the U.S. uninsured population (https://aspe.hhs.gov/sites/default/files/private/pdf/265041/trends-in-the-us- uninsured.pdf B). • Choose one condition for your discussion post about an uninsured group. (For example, In 2019, 22% of Latine individuals under age 65 were uninsured, Figure 4 p.6). For your discussion board post include all of the following: 1. Suppose six individuals, within your criteria, are randomly selected. Show or explain your thinking for each problem. (Note: This is a binomial probability from Section 5.2, they either have insurance or they do not.) A. Let having insurance be a success. State the values of your p, q, and n for your chosen condition. Explain your thinking. (For my example, if 22% do not have insurance, then that is my q (failure) value, my p=0.78, q=0.22, n-6 (6 individuals are randomly selected in this problem). B. Use excel or a T184 to determine the…
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You may need to use the appropriate appendix table or technology to answer this question. Individuals filing federal income tax returns prior to March 31 received an average refund of $1,057. Consider the population of "last-minute" filers who mail their tax return during the last five days of the income tax period (typically April 10 to April 15). For a sample of 400 individuals who filed a tax return between April 10 and 15, the sample mean refund was $910. Based on prior experience, a population standard deviation of σ = $1,600 may be assumed. What is the test statistic? (Round your answer to two decimal places.) What is the p-value? (Round your answer to four decimal places.) p-value = Find the value of the test statistic. (Round your answer to two decimal places.) State the critical values for the rejection rule. (Use ? = 0.05. Round your answer to two decimal places. If the test is one-tailed, enter NONE for the unused tail.) test statistic≤test statistic≥

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