Use a 0.05 significance level to test the claim that there is a difference between the actual and reported heights, in inches, for 12-16 year old boys. The data is listed in the table below.Click the icon to view the data table of the reported heights.Let μ₁ denote the median of the first variable and μ₂ denote the median of the second variable. What are the null and alternative hypotheses?OA. Ho: H₁ H₂H₁: H₁ H₂C. Ho: H₁ H2H₁: H₁ H₂Find the test statistic.Test statistic = -2.26 (Round to two decimal places as needed.)CHMore InfoOB. Ho: H₁ H₂H₁: P1 P₂OD. Ho: ₁2H₂H₁ H₁ H₂Actual 61.9 67.9 71.2 61.4 62.3 68.8 65.9 70.8 61.9 66.7Reported 62 68 71 62 61 69 66 70 62 67Actual 59.6 71.9 64.5 64.4 71.9 69.1 68.8 68.1 71.7 62.5Reported 60 72 64 64 72 69 68 68 72 62Actual 65.2 59.9 64.8 65.5 67Reported 65 60 65 66 67Print68.7 65.2 71.4 69.2 60.969 65 72 69 61Done- X

T test is the type of parametric test. It follows the assumption of normal distribution. It compare…

Answered: Use a 0.05 significance level to test…

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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What is the IQR for the following box-and-whisker data: Min = 154, LQ = 180, Med = 197.5, UQ = 211, Max = 265 O a. 26 O b. 43.5 O c. 111 O d. 31
The data in the accompanying table give the weights (in g) of randomly selected quarters that were minted after 1964. The quarters are supposed to have a median weight of 5.670 g. Use the sign test and a 0.01 significance level to test the claim that the median is equal to 5.670 g. Do quarters appear to be minted according to specifications? Click the icon to view the data. What are the null and alternative hypotheses? O A. Ho: Median weight = 5.670 g H₁: Median weight 5.670 g More Info OB. Ho: Median weight ≤ 5.670 g H₁: Median weight = 5.670 g O D. Ho: Median weight = 5.670 g H₁: Median weight # 5.670 g 0 Post-1964 Quarters 5.5863 5.5764 5.6582 5.5649 5.5844 5.7402 5.7367 5.6057 5.5995 5.6165 5.6448 5.5974 5.6609 5.6363 5.6352 5.5986 5.5797 5.5501 5.6238 5.6462 5.6423 5.7021 5.5481 5.7078 5.6055 5.5324 5.7417 5.6776 5.5924 5.5691 5.5666 5.7287 5.6312 5.6017 5.6626 5.6436 5.6634 5.6132 5.5441 5.6363 -
Children with autism differ significantly from children without autism on tests of varbal ability. repeated measures? or ind. with two tails?
The data found below measure the amounts of greenhouse gas emissions from three types of vehicles. The measurements are in tons per year, expressed as CO2 equivalents. Use a 0.025 significance level to test the claim that the different types of vehicle have the same mean amount of greenhouse gas emissions. Based on the results, does the type of vehicle appear to affect the amount of greenhouse gas emissions? Click the icon to view the data. What are the hypotheses for this test? O A. Ho: At least one of the means is different from the others. H₁ H₁ H₂ H3 OB. Ho: H₁ H₁: H1 H₂ =H3 H₂ H3 OC. Ho: H₁ H₂ H3 H₁: At least one of the means is different from the others. O D. Ho: H₁ H₂ H3 H₁: H₁ H₂ =H3 Data Table Type A 6.2 6.2 6.5 7.4 7.1 6.1 6.9 7.3 6.9 6.7 Print C Type B 7.7 7.5 8.7 8.1 8.6 8.8 8.4 8.5 Type C 8.5 9.3 9.6 9.1 9.1 9.1 9.8 9.5 9.9 Done O X
The data found below measure the amounts of greenhouse gas emissions from three types of vehicles. The measurements are in tons per year, expressed as CO2 equivalents. Use a 0.05 significance level to test the claim that the different types of vehicle have the same mean amount of greenhouse gas emissions. Based on the results, does the type of vehicle appear to affect the amount of greenhouse gas emissions? Click the icon to view the data. What are the hypotheses for this test? OA. Ho: H₁ H₂ H3 H₁: H₁ H₂ H3 OB. Ho: H₁ H₂ =H3 H₁: At least one of the means is different from the others. O C. Ho: At least one of the means is different from the others. H₁: H₁=H₂ =H3 OD. Ho: H₁ H₂ H3 H₁: H1₁ H₂ H3 Determine the test statistic. F = (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to two decimal places as needed.) What is the conclusion of the test? the null hypothesis. Conclude that the type of vehicle Data Table C Type A 6.5 6.2 6.5 6.5 5.8 7.5 6.4 6.4 6.2 6.9…
Fertilizer: A new type of fertilizer is being tested on a plot of land in an orange grove, to see whether it increases the amount of fruit produced. The mean number of pounds of fruit on this plot of land with the old fertilizer was 381 pounds. Agriculture scientists believe that the new fertilizer may increase the yield. State the appropriate null and alternate hypotheses. Answer the following question.
Data Table Type A 7.3 7.6 7.1 7.1 7.1 7.9 6.4. 6.9 7.5 7.2 Type B 8.7 8.2 8.3 7.7 8.8 8.2 8.7 7.1 Type С 8.5 8.3 9.3 9.2 9.3 9.5 8.8 8.7 8.5 - X
Use the following R output to answer the question. Use the following R output to answer the question. >chisq.test(data) Pearsons Chi-square test data:data X-squared=4.7194, df=2, p-value=0.09445 If the significance level is 5%, what conclusion can you make? Group of answer choices A. fail to reject the alternative hypothesis B. cannot determine C. fail to reject the null hypothesis D. reject the null hypothesis and accept the alternative hypothesis
Is the mean length of adult invasive Lionfish in the Atlantic coast of South Carolina greater than 355.6mm? (don't answer this, this is just so you know what the data is for) Answer the following questions and use the provided excel images, data on the far left is lionfish data Question 2. Describe your null and alternative hypotheses using symbols. Question 3. Select the significance level (a). Question 4. Select an appropriate test, calculate the test statistic and p-value using Excel. Question 5. Determine whether you should reject or not reject the null hypothesis based on the p-value. State the reason to support your answer. Question 6. Describe your conclusion in words in the context of your research question. paste data below Use this sheet to perform a 1-sample t-test. 457.2 431.9 COUNT MEAN 20.00 437.45 438.2 STANDARD DEVIATION 16.48 436.4 STANDARD ERROR 3.69 416.8 454.6 Confidence Interval 95% 456.6 Level of Significance 0.05 453.1 404.9 447.9 437.9 HYPOTHESIS TEST t-VALUE…
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