## Conducting a Hypothesis Test for Categorical Data### ScenarioA person randomly selected 100 checks and recorded the cents portions of those checks. The table below lists those cents portions categorized according to the indicated values. Using a 0.025 significance level, we aim to test the claim that the four categories are equally likely. The person expected that many checks for whole dollar amounts would result in a disproportionately high frequency for the first category. Do the results support that expectation?### Data Table#### Cents Portion of Check| Cents Portion of Check | Number ||------------------------|--------|| 0-24 | 58 || 25-49 | 11 || 50-74 | 17 || 75-99 | 14 |### Hypothesis Test1. **Calculate the Test Statistic**: - The test statistic is calculated using the chi-square goodness of fit test. - The observed and expected frequencies are required for this calculation. - Test statistic formula: \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\)2. **Determine the Critical Value**: - Use the chi-square distribution table to find the critical value at a 0.025 significance level.3. **State the Conclusion**: - Compare the test statistic to the critical value to determine whether to reject the null hypothesis.### Form Fields:- **Test Statistic**: (Round to three decimal places as needed.)- **Critical Value**: (Round to three decimal places as needed.)### Conclusion Selection:Choose the appropriate conclusion based on the comparison between the test statistic and the critical value:- Drop-down options: - "Sufficient evidence" or "Insufficient evidence" - "Reject" or "Fail to reject"### Conclusion Statement:- If the test statistic exceeds the critical value, there is sufficient evidence to warrant rejection of the claim that the four categories are equally likely. The results support the expectation that the frequency for the first category is disproportionately high.- Conversely, if the test statistic does not exceed the critical value, the results do not support the expectation that the frequency for the first category is disproportionately high.Remember to click the link provided to view the chi-square distribution table for critical values. ### Chi-square Distribution TableThe Chi-square distribution table is used to find the critical value of the Chi-square statistic given the degrees of freedom and the desired significance level (alpha). This table assists in hypothesis testing, particularly in tests of independence and goodness of fit.**Table Explanation:**The table presents the critical values for different degrees of freedom (rows) and areas to the right of the critical value (columns). For instance, the critical value for 3 degrees of freedom with an area of 0.05 to the right is 7.815. It is organized in a grid:- **Rows:** Representing the degrees of freedom (from 1 to 10).- **Columns:** Providing the critical values for various right-tail probabilities (significance levels).**Area to the Right of the Critical Value:**- **Degrees of Freedom | 0.995 | 0.99 | 0.975 | 0.95 | 0.90 | 0.10 | 0.05 | 0.025 | 0.01 | 0.005**- **1** | - | - | 0.001 | 0.004 | 0.016 | 2.706 | 3.841 | 5.024 | 6.635 | 7.879- **2** | 0.010 | 0.020 | 0.051 | 0.103 | 0.211 | 4.605 | 5.991 | 7.378 | 9.210 | 10.597- **3** | 0.072 | 0.115 | 0.216 | 0.352 | 0.584 | 6.251 | 7.815 | 9.348 | 11.345 | 12.838- **4** | 0.207 | 0.297 | 0.484 | 0.711 | 1.064 | 7.779 | 9.488 | 11.143 | 13.277 | 14.860- **5** | 0.412 | 0.554 | 0.831 | 1.145 | 1.610 | 9.236 | 11.071 | 12.833 | 15.086 | 16.750- **6** | 0.

Given information: The sample size is n=100. The level of significance is α = 0.025. It is asked to…

Answered: lues. Use a 0.025 significance level to…

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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17)

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