Customers at 3 resorts were surveyed about customer satisfaction by asking wheth
ID: 3223226 • Letter: C
Question
Customers at 3 resorts were surveyed about customer satisfaction by asking whether they plan to return or would recommend the resort to a friend. Based on the data The test statistic 5.012 is less than the critical value 5.991, pvalue = 0.07, H_0: Quality of resorts is equal The test statistic 26.187 exceeds the critical value 5.994, pvalue = 0.01, reject H_0: Quality of resorts is equal, conclude quality of resorts is unequal The test statistic 26.187 exceeds the critical value 5.991, pvalue = 0.00, reject H_0: Quality of resorts is equal, conclude quality of resorts is unequal None of the above.Explanation / 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 X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.991
since our test is right tailed,reject Ho when ^2 o > 5.991
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 26.185
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ^2| =26.185 & | ^2 | =5.991
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
ANSWERS
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null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 26.185
critical value: 5.991
p-value:0
decision: reject Ho
The test statistic 26.187 exceeds me criticaL va’ue 5991, pvalue 000, reJject Ho...
col1 col2 col3 row 1 163 143 79 385 row 2 7 4 18 29 TOTALS 170 147 97 N = 414Related Questions
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