Use a significance level of0.05 to test the claim.One-way ANOVA:SourceDFSSMSSize3010.001.60.264Error506.25Total1180Null Hypothesis, HOAlternative Hypothesis, H1Test Statistic, FP-valueSignificance levelConclusion

One-way ANOVA : Source DF SS MS F P Size 3 30 10 1.6 0.264 Error 8 50 6.25 Total 11 80 ___…

Answered: Use a significance level of 0.05 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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The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n=761 and the test statistic is t=55.668. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10. State the null and alternative hypotheses. H₁> 2.84 2.84 (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed.)
Using p value method only, with significance level of 0.05 What is H0? What is H1? What is P value?
The p-value for a t-statistic was .008. What is the best way to report the level of significance? p<.008 p<.05 p<.01 p<.001
For the following summary table for a one-way ANOVA, fill in the missing items (indicated by asterisks), identify the null and alternative hypotheses, then use the 0.05 level of significance in reaching a conclusion regarding the null hypothesis. Variation Source Sum of Squares Degrees of Freedom Mean Square F Treatments 6752.0 2 3376.0 *** Error 30178.0 *** *** Total 36930.0 29
I need help with this problem Conduct a hypothesis test about the mean cost per square foot at a chosen significance level. Reference an internet source for the current average (or median) cost per square foot of homes in Orange County. Discuss/explain your choice of significance level for your hypothesis test in #1 as it relates to the power of a test and Type I/II errors.
Given the following ANOVA table: Degrees Mean Source of of Variance Sum of Sum of F-stat Squares Freedom Squares | Regression 2 5.5 Residual 27 2.507 Total What is the value of SST? Please report your answer in 3 decimal places.
Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less f than 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Click the icon to view the display from technology. @ 2 View an example Get more help. W S Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. Ho: =4.00 Mbps H₁: 4.00 Mbps मे Identify the test statistic. (Round to two decimal places as needed.) X # 3 E 80 F3 D C $ 4 Q 20 F4 R F V % 5 FS T G A 6 B MacBook Air F6 Y H & 7 N 8 F7 U C J * 8 M OB. Ho: F10 . : - ; I option Clear all { [ ? 1 + = 11 Check answer F12 } 1 delete return shif C
Big babies: The National Health Statistics Reports described a study in which a sample of 335 one- year-old baby boys were weighed. Their mean weight was 25.2 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the a= 0.01 level of significance and the critical value method with the Critical Values for the Student's t Distribution Table. Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0 μ-25 H1: u#25 This hypothesis test is a two-tailed V test. Part 2 of 5 Find the critical value(s). Round the answer(s) to three decimal places. If there is more than one critical value, separate them with commas. Critical value(s): 2.576 , 2.576 Part: 2/5 Part 3 of 5 (b) Compute the value of the test statistic. Round the answer to at least three decimal places.
Think about doing a Repeated-Measures ANOVA, Step 2. Below is an ANOVA Source Table that is only partially filled out (just the SS and df columns are filled out). From this information, what are the F-obtained and Eta-squared? SOURCE SS df MS Fobt Between 8.00 2 Within 10.00 6 Participants 4.67 2 Error 5.33 4 Total 18.00 8 Group of answer choices F-obtained = 3.00, Eta-squared = 0.15 F-obtained = 3.00, Eta-squared = 0.60 F-obtained = 8.00, Eta-squared = 0.40 F-obtained = 8.00, Eta-squared = 0.80
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Use a significance level of 0.05 to test the claim. One-way ANOVA: Source Size Error Total DF SS 3 8 11 MS 10.00 6.25 F 1.6 30 0.264 50 80 Null Hypothesis, HO Alternative Hypothesis, H1 Test Statistic, F P-value Significance level Conclusion