00:01
Hello students, here we have given the sample data and by using that we want to conduct the goodness of fit test to see whether the following sample appears to have been selected from the normal probability distribution.
00:15
That means we have the null hypothesis that is the population has normal distribution and the alternative hypothesis is the population does not have the normal distribution.
00:26
Now here we have given the sample mean x bar is equal to 71 .4 sample standard deviation s is equal to 19.
00:37
Then now here we are going to use the goodness of fit test.
00:42
So the test statistic is given by chi square is equal to summation of observed frequency minus expected bracket square divided by expected frequency.
00:53
Now suppose this is equation number one.
00:56
So for that first we need to compute the observed frequency and the expected.
01:02
Now here we are going to divide the data into the five classes.
01:07
So first column is for the classes in that the first class is from 0 to 20 second is from 20 to 40 then 40 to 60 60 to 80 and the last one is 80 to 100.
01:28
Now in the first class that is between 0 to 20 the frequency is 0 that is there is no any observation that belongs to this class.
01:39
Then the frequency of the second class is 1 then this is 10 this one is 4 and this again 10.
01:49
So the total frequency is 25 and now this frequency is nothing but the observed frequency that is oi.
01:56
Then to calculate the expected frequency the formula for the expected frequency ei is equal to the total of the observed frequency which is 25 divided by the number of classes which is 5.
02:11
So we will get this is equal to 5.
02:15
So the expected frequency is 5 for each of the classes.
02:18
And now by using this we can calculate the test statistic value.
02:24
So from 1, chi square is equal to 0 minus 5 bracket square divided by 5 plus 1 minus 5 bracket square divided by 5 plus 10 minus 5 bracket square divided by 5 plus 4 minus 5 bracket square divided by 5 and plus 10 minus 5 bracket square divided by 5.
02:52
So therefore the value of chi square is equal to 13 .4.
02:59
Then the degrees of freedom is equal to k minus 1 where k is the number of classes.
03:06
So this is 5 minus 1 is equal to 4 and we will get the required p value is equal to 0 .0095...