00:01
In this question we are giving a data as 36, 52, 63, 68, 35, 59, 74, 31, 31, 37, 31, 37, 42, 51, 51, 58, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48.
00:33
53 53 68 and 42 in this we are given n is equals to 20 alpha is equal to 0 .05 and it is said that median is equals to 53 so we write the hypothesis for this null hypothesis is median is equal to 53 and alternative hypothesis is median is not equals to 53.
01:11
So from here we find d .i.
01:15
And the formula for d .i.
01:16
Is equals to xi minus median.
01:20
Here we have to find the critical value.
01:23
So if we see that the signs for this d .i.
01:27
36 minus 53 is negative we get a negative sign for this we also get a negative sign for 63 minus 53 positive sign 68 minus 53 positive sign for this we get negative sign positive here we get positive sign next we get negative sign negative sign and negative sign 51 minus 53 is negative then we get positive sign positive negative negative negative negative 0, 0 63.
02:01
For 63, this is positive, positive and negative...