RaregjaphStylesTTest Results: tstas -1.745, p-value=0.9514) What is the P-value? View the video Hypothesis Testing for the Mean (Sigma Unknown)Part Il to learn how to use Google Sheets to calculate the P-value of a t-test (transcriptfor Hypothesis Testing for the Mean (Sigma Unknown) Part Ii video).030p-value = ?20I.J) Make a decision to reject or fail to reject the null hypothesis using either the teststatistic or the P-value. Note that the same conclusion will be reached using eithermethod..100.004.0 -3.5 -3D 25 -20 -15 -10 -0s 0.0 0.51015 20 253.0 3.5 4.0K) Interpret the decision in the context of the original claim.G) Calculate the test statistic. Replace the ?s in the formula with the appropriate values.Round to the nearest hundredth.え一ドt 3=L) If you lower the level of significance to a = 0.01, does your decision change? Explainyour reasoning.?-?t =M) Describe the type I and type II errors that could occur in our test by completing each ofthe following statements.Atuition cost ishypothesis He: u ?$10,230.(type I, type II) error will occur when the actual mean of college(at most, at least) $10,230 but you reject the nullt = ?H) Add the test statistic to your graph from part F, and insert the image below.(type I, type II) error will occur when the actusl mean of college(less than, greater than) $10,230 but you fail totuition cost isreject the null hypothesis, Hei u 7$10,230.約to search# e Upload Asignmen.Answered: E) Using.Decumenti - WerdPractice Werk.deGKO5. Calender - iregal. ILEIFontParagraphStylesMean 8953Standard Deviation 3273.46According to the College Board's website, the average tuition cost per year for a public four-yearcollege (in-state cost) is $10,230.A) Assume that you disagree with the claim. Do you think the average tuition cost is higher?Or lower?E) Using a significance level a = 0.05, find the critical value using the following steps.1. What are the degrees of freedom?Average tuition cost is higher than $10230.B) Using your answer in part A, write a set of hypotheses (null hypothesis and alternativehypothesis) to represent that claim and its complement. Replace the 7s below with theappropriate inequality signs (<, S,>,2). Highlight the hypothesis that represents yourclaim.d.f.=n-1(1)Here the sample size is 20, we know degrees of freedom =n- 1=20-1 19Ho:p?$10,2302. Will this be a one-tailed or two tailed test?Ho:u=10230 Hi:p>10230Ha-u ? $10,230A right-tailed test (one-tailed), for which a t-test for one mean will be used.C) Gather data for tuition cost of 20 different universities. Assume the data you gather isnormally distributed.3. Using table 5 from the back of your textbook (page A18 in Appendix B), what isthe critical value?How to Create your data set:Information provided, the significance level is a=0.05\alpha =0.05, and the criticalvalue for a right-tailed test is t1.729Follow the directions to fill in the table below with 20 college's tuition costs.Navigate to https://www.collegetuitioncompare.com/compare/tables/In the drop down menus, ONLY select School Type as "Public School and SchoolLevel as "4 year or higher." Leave the rest of the drop down menus as is. Then select"update."Scroll down to the bottom of the list and for "Number of schools to show," select "AlLThe list given will be in alphabetical order. You will be selecting 20 college tuitioncosts from this list. Do NOT just use the first 20 entries or even 20 consecutiveentries. You will need to randomly select 20 entries from the entire list. Try to makesure you have a good range of tuition costs as well.Write the tuition cost from the first column, "Tuition & Fees / In-State" for that collegein the table below.F) Add the critical values to the graph below and shade the rejection region(Special Instruction: For this part, it is recommended that you do this by hand You can printout this page and use the image below as a template to write on or you can create a handwrittennormal curve. Take a photo of your graph and replace the image below with it. If you need anytechnical assistance with this process.Since it is observed that t=-1.745

Name: RaregjaphStylesTTest Results: tstas -1.745, p-value=0.9514) What is t
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Description: RaregjaphStylesTTest Results: tstas -1.745, p-value=0.9514) What is the P-value? View the video Hypothesis Testing for the Mean (Sigma Unknown)Part Il to learn how to use Google Sheets to calculate the P-value of a t-test (transcriptfor Hypothesis T