00:01
Hello students, let's solve this problem.
00:02
The problem says that the research found the influence of the background noise on the class performance of the children.
00:10
They found that the claiming music lead to better performance on a math tax compared to a non -music condition.
00:20
A research selected one class of n is equal to 18 students who listened the claiming music.
00:35
A second class, n is equal to 18, serves as a controlled group with no music.
00:44
The average of the student in the first class is m equal to 86 .4 and with ss is equal to 1550.
01:03
Compare with an average of m is equal to 78 .8 with ss is equal to 1204.
01:17
For the student in the second class, the population are normally distributed and use alpha is equal to 0 .05.
01:31
So the first question is given is there a significant difference between the two music condition that is is given so we have to solve this so first of all we find out that is of s one because n1 we can write down that is n1 is equal to 18 so we find out s1 that is under root of s multiplied with s1 divided by n1 minus 1.
02:15
So under root of 1550 divided by of 17, so that we get 9 .586.
02:30
Next we find out this of s2 and this one, and this one, we write down that this part we are just write down that is of s when you find out s 2 under root of s s divided by n1 minus of 1 so in the similar process second s is given 1 to 04 divided by 17 so we get 8 .4 and x bar that is m mean is 86 .4 similarly y bar m m 78 .8 now for the s we have to find out so the formula we can write down s is equal to under root n1 minus 1 s 1 square plus n2 minus 1, n2 minus 1 multiplied with s 2 square divided by n1 plus n2 minus 2.
04:12
So we have to put all those values that is root under 17 multiplied with 9 .5.
04:25
86 square plus 17 multiplied with 8 .4152 divided by n1 plus n2 minus of 2.
04:52
So after doing this calculation s that is 8 .9999 this week.
05:01
So the approximate value that is equal to 9.
05:07
Here we want to test that h0, mu1 is equal to mu2, versus h1, mu, 1, not equal to mu2.
05:29
So the test statistics, so the formula of the test statistics, t is equal to x2, x bar minus y bar divided by s under root of 1 by n1 one divided by n1 and one divided by of n2 so we have to put all those values that is 86 .4 minus 78 .8 divided by 9 divided by 9 under root of 1 by 18, 1 divided by 18 and 1 divided by 18.
06:16
So we get t is equal to 2 .5333.
06:25
We can say this...