### Chi-Square Test of Independence: Flu Shot and Flu IncidenceYou are conducting a test of independence to examine if getting the flu is independent of whether or not you got a flu shot. Use \(\alpha = 0.05\). Report all answers accurate to four decimal places.Here is the sample data:| Category | Got the Flu | Did not get the Flu ||---------------------|-------------|---------------------|| Got Flu Shot | 18 | 118 || Did not get Flu Shot| 51 | 19 |#### Expected ObservationsThe expected observations for this table would be:| Category | Got the Flu | Did not get the Flu ||---------------------|-------------|---------------------|| Got Flu Shot | | || Did not get Flu Shot| | |#### Chi-Square CalculationThe resulting \(\frac{(O - E)^2}{E}\) for finding the test are:| Category | Got the Flu | Did not get the Flu ||---------------------|-------------|---------------------|| Got Flu Shot | | || Did not get Flu Shot| | |#### Chi-Square Test StatisticWhat is the chi-square test-statistic for this data? (add the results from the previous table)\[\chi^2 = \]#### P-Value CalculationWhat is the p-value for this test-statistic?P-Value = On this educational webpage, detailed explanations and step-by-step calculations will follow this initial setup to guide you through each stage of conducting a chi-square test of independence.

CategoryGot the fluDid not get the fluGot Flu Shot18118Did not get Flu Shot5119

Answered: You are conducting a test of…

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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This is a Statistics problem. Please code in R.
Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.10 to test for a difference between the weights of discarded paper (in pounds) and weights of discarded plastic (in pounds). Click the icon to view the data In this example, Hg is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis test? O A. H9: Ha =0 H1: Ha 0
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**Frequency distribution table** The city council was concerned that allowing more bars within city limits would increase the number of drunk drivers on city streets. A researcher was asked to statistically determine how many bar patrons exceed the legal limit for driving when leaving a bar. The researcher asked 20 bar patrons to take a test to determine their blood alcohol levels as they were leaving the bar. The scores are shown below. 0.05 0.08 0.09 0.12 0.03 0.07 0.09 0.12 0.10 0.06 0.10 0.02 0.08 0.00 0.07 0.00 0.14 0.02 0.11 0.10
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.)
I 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 The table available below shows the weights (in pounds) for a sample from the others? E Click the icon to view the vacuum cleaner weight data. Let Hgu. PBLU and PTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? O A. Ho: Heu * PBLU HTC H PBU PBLU =HTC Weight of Vacuum Cleaners by Type O B. Ho: At least one of the means is different. H PBu = HBLUu =PTC &c. Họ: PBU = HBLU =HTC Bagged upright 22 19 Bagless upright Top canister 21 22 23 25 H: At least one of the means different. 17 20 15 18 18 19 O D. Ho: PBU = PBLU "HTC 23 20 26 25 22| 23 Ha: Peu * PBLU *HTC What is the test statistic? Print Done F= (Round to two decimal places as needed.)
Please see the picture below. I only am given one attempt at this problem.
ed in in the table 2. Find the mean, median, and mode for the following sample of scores: one person the sample 12. A sample One persoi 5 4 5 2 7 1 3 5 ust use the s found stribution nidpoint of students edian, you ates the highest quency. epresen-

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