00:01
So here the data for new jersey is given.
00:06
It is given the sample of size 100 is considered and the sample mean, let this be y bar is given as 558 and the standard deviation of y is obtained as 8.
00:20
Now first we were asked to find a 95 percentage confidence interval for mean score.
00:32
Now, standard deviation population, standard deviation of the data, sigma is not known.
00:37
So we can use t tables to find the confidence interval, which is 5x or minus t alpha by 2 with the degrees of freedom n minus 1 and s by root n.
00:48
This will give you the confidence interval.
00:51
And t alpha by 2, n minus 1 means here alpha will be 0 .05 since we have 95 percentage confidence.
01:01
And n minus 1 will be 100 minus 1.
01:04
And using excel table, the formula is shown here, the answer is 1 .9842.
01:11
Now, substituting all the values in the confidence interval, we have 58 plus or minus 1 .9842 into 8 by root 100.
01:23
And on calculation, the confidence interval is 504 .85032.
01:31
To 61 .1497.
01:36
So this is the 95 percentage confidence in the 4 for the means scores for new jersey.
01:46
Now for eova, a data is also given which is given that n equal to 200 and x bar is equal to 62 xx is equal to 11.
02:00
Here we need to construct a 90 percentage confidence interval for difference in means between e .o .o .a.
02:19
And new jersey.
02:21
So we have to construct a confidence interval for the difference of the means of these two.
02:28
So the required equation is x bar minus y bar plus or minus t alpha by 2 with a degrees of freedom n1 plus n2 minus 2 into here the expression is n1 minus 1 into s 1 square plus n2 2 2 square divided by n 1 plus n 2 minus 2 this is the pooled variance and here you have to multiply to get the standard error 1 by n 1 plus 1 by n 2 then take the square root of all this term will give you the standard error of the difference between these two variables.
03:14
Now, substituting the values, we have 62 minus 58, plus or minus, the t value is here 1 .65...