In: Statistics and Probability
The following table contains observed frequencies for a sample of 200.
Test for independence of the row and column variables using a=.05. Compute the value of the test statistic (to 2 decimals). The -value is - Select your answer -less than .005between .005 and .01between .01 and .025between .025 and .05between .05 and .10greater than .10 What is your conclusion? - Select your answer -Cannot reject the assumption that rows and columns are independent Conclude the row variable and column variable are not independent |
the necessary calculation table be:-
row variable | A | B | C | total | |
P | observed frequency(Oi) | 40 | 48 | 45 | 133 |
expected frequency(Ei) |
(50*133)/200 = 33.25 |
46.55 | 53.20 | ||
1.3703 | 0.0452 | 1.2639 | |||
Q | observed frequency(Oi) | 10 | 22 | 35 | 67 |
expected frequency(Ei) | 16.75 | 23.45 | 26.80 | ||
2.7201 | 0.0897 | 2.5090 | |||
column total of observed frequency | 50 | 70 | 80 |
hypothesis:-
the row and column variables are independent.
the row and column variables are not independent.
the test statistic be:-
The p-value is :- between 0.01 and 0.025
[ df = (3-1)*(2-1)= 2
fro chi square table for df =2,chi square = 8.00 ]
my conclusion is:-
Conclude the row variable and column variable are not independent.
[ p value < 0.025 < 0.05 (alpha).
so we reject our null hypothesis.]
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