The table below cross-classifies marital status and happiness. The values in par
ID: 3250370 • Letter: T
Question
The table below cross-classifies marital status and happiness. The values in parentheses are standardized residuals. 1) Software reports that the 2= 236.4. Find the p-value. If = 0.05, would you conclude that marital status and happiness are statistically independent? Be sure to pay attention to the degrees of freedom in this table when calculating the p-value.
Show work please.
The table below cross-classifies marital status and happiness. The values in parentheses are standardized residuals. MARITAL HAPPINESS Pretty Happy STATUS Very Happy Not Too Happy Total 93 (-10.00 720 (-5.4) Married 600 (13.1) 1413 63 (-2.2) 142 (-0.2) 51 (3.4) Widowed 256 304 (3.2) 485 93 (-6.1) 88 (3.6) Divorced Separated 19 (-2.7) 51 (-1.2) 31 (5.3) 101 730 Never Married 144 (-7.4) 459 (4.2) 127 (4.0) 919 390 Total 1676 2985 Source: 2006 GSS 1) Software reports that the 236.4. Find the p-value. If a -0.05, would you conclude that marital status and happiness are statistically independent? Be sure to pay attention to the degrees of freedom in this table when calculating the p-value.Explanation / Answer
Answer to the question)
Given that chi square vlaue of the statistic is : 236.4
The degree of freedom is calculated by the formula
df = (r -1) * (c - 1)
where :
r = total number of rows
c = total number of columns
.
In the table above (exclude the last row fo total) we got r = 5
c = 3
.
On plugging these vlaues we get:
df = (5-1) * (3-1)
d f= 8
.
Now we got df = 8 , adn chi square statistic = 236.4 ,w e can find the P value with the help of excel or online P value calculator
.
With excel, the formula is :
=1-CHISQ.DIST(236.4,8,1)
we get P value = 0
.
On online calculator as well it gives the result : P value < 0.00001
.
Thus as significance level 0.05
We conclude that as P value is less than 0.05 , we reject the null hypothesis and conclude that the marital stauts and happiness are statstically significant.
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