00:01
Given in the question, given n is equal to 20, mean x bar is equal to 0 .9, and standard deviation to note of a sigma is equal to 0 .5.
00:16
And first, the 90 % confidence interval, the 90 % confidence of terra.
00:29
For mean, muil is calculated as x bar plus minus test statistic values.
00:37
At alpha over 2 significant sluil and n minus 1 degrees of freedom times of standard deviation over square root of n now on substituting these values here we get xb0 .9 plus minus t alpha over 2 that is t at 0 .05 significance level and 20 minus 1 is 19 so 19 degrees of freedom times of sigma value 0 .0 5.
01:09
0 .5 over square root of 20.
01:12
Now on calculating this we get 0 .9 plus minus 1 .7291 times of 0 .118, which is equal to 0 .9 plus minus 0 .204.
01:29
Therefore we have at 90 % confidence interwe.
01:33
New value lies between 0 .696 and 1 .104.
01:42
Therefore, this is the required conference interval.
01:47
Width will be equal to 2 times of 0 .204, that is 0 .408.
01:59
Now from this, coming to the first part of the question, so this is the solution for the first part of the question...