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QUESTION 13 3 points Save Answer t -= . and df = degrees of freedom» n- There is

ID: 3353201 • Letter: Q

Question

QUESTION 13 3 points Save Answer t -= . and df = degrees of freedom» n- There is one major issue purposely avoided in quiz 5 and now so far in this quit. The issues ane o What is t7 What is z e When to use 1? When to use z? You come across t n CD 6.5, CO 82·CD94,CD96 Both the formulas look ake except that one has s and one has . two different symbols tor andard devation. · df-degrees of freedom, sormething special for t Both t and z behave alke when n is large like n more than 30 A QUICK HINT Use t i# you see that population standand deviantion is not given or only sample standard deviation (a) is given tabiens -one tal and two tal Unilike tive z tables (0 to z etc and critical table), you wislonly see two t . Also reading these two t tables is quite different than reading z tables. . Click both the tables -see the . In onetail table, area to the number 2.776-that appears in both the tables RIGHT of 2.776 is 0.02S or 2.5% and (oaivalenty, area to In your Co 9.6, it will ask you to find a eriftical number (one tail or tero tail) using t table with df - 4 and level of significance 0.06, then you can ind it two ways use har of level of . If it ks you to find two-tal number, you must go totwa tai table-you gotoone tai table, you must significance Also n CD a 4, naaka you cor tail table truct 95% onfidence interval, then you go to ODS courm of two tail or 0.025 column of one e Here is a typical probiem from CD 8 The tollowing measurements in picocuries per liter ) were recorded by a set of carbon dioskde detectors installed in a manufacturing faclity 700.2, 784.3,803 8, B06.8. 780.6, 7948

Explanation / Answer

GIven that 799.2,784.3,803.8,806.8,780.5,794.8

=> Degrees freedom = n-1 = 6-1 = 5

=> 95% critical number = 2.571

=> sample mean = 794.9

=> s = 10.6

=> Up CI number = X + t * s/sqrt(n) = 794.9 + 2.571*(10.6/sqrt(6)) = 806.0

=> Low CI number = 794.9 - 2.571*(10.6/sqrt(6)) = 783.8

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