00:01
Here, the null hypothesis is the age distribution of the general canadian population fits the age distribution of the residence of red lake village.
00:54
Alternate hypothesis is the age distribution of the general canada.
01:16
Canadian population does not fit the age distribution of the residence of red lake village.
01:55
Here the level of significance given in the problem is alpha is equal to 0 .1.
02:05
Next we need to calculate the test value.
02:10
The formula is kaisal.
02:11
Equal to summation o minus e the old square divided by e in the problem we have n is equal to 448 the expected frequency is calculated as n into the probability value that is n p the calculation is shown in the table before the expected frequency is calculated for all the categories and 0 minus e the whole square divided by e is also calculated.
02:53
Therefore, kai square value is equal to summation o minus e the whole square divided by e, which is equal to 6 .61878.
03:06
Next, we calculate the p value.
03:10
Here, the degrees of freedom is equal to n minus 1, which is the number of categories, that is 4 minus 1, which is equal to 3...