**Analysis of Hospital Admissions for Motor Vehicle Crashes****Context:**Researchers conducted a study to determine if the occurrence of motor vehicle crashes, leading to hospital admissions, changes when the 13th day of a month falls on a Friday. They compared data for hospital admissions from crashes on Fridays that were the 6th of the month to those on Fridays that were the 13th. **Data Collected:**- **Friday the 6th:** - Admissions: 9, 6, 12, 12, 4, 5- **Friday the 13th:** - Admissions: 13, 12, 14, 11, 6, 12**Objective:**To assess if there is a significant difference in the number of admissions between these two specific days using a significance level of 0.05.**Steps to Undertake:**1. **Calculate the Test Statistic:** - Determine the value of the test statistic `t` and round it to three decimal places.2. **Identify Critical Values:** - Choose either: - **Option A:** The critical value is a single value `t = [ ]`. - **Option B:** The critical values are `t = ± [ ]`.3. **Interpret Results:** - Use the calculated test statistic and critical value(s) to draw a conclusion about the hypothesis that the day (Friday the 13th vs. Friday the 6th) does not affect the number of hospital admissions from motor vehicle crashes.This analysis will enable us to understand if superstitions surrounding Friday the 13th have any grounding in real-world data pertaining to vehicle crashes. Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and results are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Use a 0.05 significance level to test the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.- Friday the 6th: 9, 6, 12, 12, 4, 5- Friday the 13th: 13, 12, 14, 11, 6, 12＿B＿ The critical values are t = ± [space for value]State the result of the test. Choose the correct answer below.- ○ A. There is sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions do not appear to be affected.- ○ B. There is sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected.- ○ C. There is not sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions do not appear to be affected.- ○ D. There is not sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected.

Method: two samples independent t test statistic

Answered: Researchers collected data on the…

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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The variable ( TV channel preference) is O a. Qualititativ variable b. None of the above O c. Quantitative variable
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A random sample of 10,000 bottles of cola was taken to see whether the mean weight was 16 fluid ounces, as marked on the container. The null hypothesis is that the population mean is 16 ounces. Use the read-out below to test the hypothesis that the colas do not have a population mean of 16 fluid ounces. You do not have to do any calculations; just interpret the given data clearly and thoroughly in the context of the problem. | One-Sample T: ounces Test of u = 16 vs = 16 Variable Mean StDev SE Mean 95 CI T P ounces 10000 15.9988 0.0985 0.0010 (15.9969, 16.0007) -1.21 0.225
The table available below shows the weights (in pounds) for a sample of vacuum cleaners. The weights are classified according to vacuum cleaner type. At a = 0.10, can you conclude that at least one mean vacuum cleaner weight is different from the others? Click the icon to view the vacuum cleaner weight data. Let μBU, BLU, and μTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? A. Ho: At least one of the means is different. Ha: MBUHBLUPTC B. Ho: "BU HBLU HTC # Ha: MBUHBLUPTC C. Ho: MBUHBLU = HTC Ha: At least one of the means is different. D. Ho: MBU - HBLU = μTC Ha: HBU #BLU HTC What is the test statistic? F = 12.56 (Round to two decimal places as needed.) What is the P-value? P-value = 0.000624 (Round to three decimal places as needed.)
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In the formula for t, the "hypothesized value" is what you expect the t to be. the difference between population means. the significance level.
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130 adults with gum disease were asked the number of times per week they used to floss before their diagnoses. The (incomplete) results are shown below: of times floss per week Frequency Relative Frequency Cumulative Frequency 0. 17 0.1308 17 1 17 0.1308 34 2. 13 0.1 47 19 0.1462 66 11 77 17 0.1308 94 6. 17 0.1308 111 0.1462 7. a. Complete the table (Use 4 decimal places when applicable) b. What is the cumulative relative frequency for flossing 1 time per week? 40 Submit Question elf SIT Coudy 4.

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