00:01
Hello students, as per the given question, we need to find the 99 % confidence interval for the population mean mu.
00:08
Based on the given sample statistics, we use the following formula, which is confidence interval is equals to sample mean plus or minus critical value into standard error.
00:25
So, we have the critical value is obtained from the t -distribution, which is n minus 1 degrees of freedom and the standard error is calculated from the sample standard deviation sigma divided by the square root of sample size.
00:41
So, wherein in the first step, let us calculate standard error first, which is given by standard deviation by root of sample size.
00:51
So, after substituting the values, which is 20 .9 by under root of 484, it gives 0 .95 as the standard error when rounded to two decimals.
01:05
So, in the next step, let us find the critical value of a 99 .9 % confidence interval.
01:15
Since the sample size is relatively large, where n is equals to 484, we can calculate the z distribution instead of t -distribution for practically purpose, that is because sample size is relatively large.
01:30
So, for this, the z value is approximately equals to 3 .291, where it is rounded to three decimals...