ditieu x value)2 (Questions 79) A survey of property owners\' opinions about a s
ID: 3326654 • Letter: D
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ditieu x value)2 (Questions 79) A survey of property owners' opinions about a strect-widening pronoml slectd ample of owners' opinions were related to the distance between their home and the stree property owners was contacted and the results are shown next. was taken to determine i Front Footage ior Indecided Against Opinion Under 45 foet _12 45-1 20 fect 35 30 Over 120 feet 3 7. What is the fre frontage? Show your solution A) 1.1 B) 3.9 C), 5.0 D) 5.5 E) Cannot decide e expected frequency for neonle against the proiect and who havc over 120 feet of property foot- R/ 8. What is the critical value at the 10% level of significance? A) 7.779 B) 9.236 C) 9.448 D) 11.070 E) 1.064 9. How many degrees of frcedom are there? A) 3 B) 4 C) 5 D) 6 E) 9Explanation / Answer
Given table data is as below
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calculation formula for E table matrix
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expected frequecies calculated by applying E - table matrix formulae
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
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set up null vs alternative as
null, Ho: no relation b/w opinion and front footage
alternative, H1: exists a relation b/w opinion and front footage
level of significance, = 0.1
from standard normal table, chi square value at right tailed, ^2 /2 =7.779
since our test is right tailed,reject Ho when ^2 o > 7.779
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 6.784
critical value
the value of |^2 | at los 0.1 with d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4 is 7.779
we got | ^2| =6.784 & | ^2 | =7.779
make decision
hence value of | ^2 o | < | ^2 | and here we do not reject Ho
^2 p_value =0.148
ANSWERS
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Q7.
3.9
Q8.
critical value: 7.779
Q9.
d.f (r-1)(c-1)= ( 3 -1 ) * ( 3 - 1 ) = 2 * 2 = 4null, Ho: no relation b/w opinion and front footage
alternative, H1: exists a relation b/w opinion and front footage
test statistic: 6.784
critical value: 9.488
p-value:0.148
decision: do not reject Ho
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