00:01
So in the given question, we are told that we want to estimate the mean time college students spent watching online videos each day.
00:09
So the estimate must be within 0 .25 minutes of the population mean.
00:15
And what we are told is to determine the required sample size to construct 97 percentage confidence interval.
00:32
And confidence interval for the population mean.
00:35
So we are told to assume the population standard deviation to be equal to 1 .3 minutes and we are told a condition that the estimate must be within 0 .25 minutes of the population mean.
00:51
So what we have to find first is the required sample size, right? so what we can do over here is the formula for finding the confidence interval.
01:02
Is x bar which is the mean plus or minus the critical value of the test statistic which we use the z value over here times the standard deviation divided by the square root of the sample size so this is what we used to find the confidence interval right and the critical value of z is what changes over here corresponding to what percentage of confidence interval we are taking so over here the critical value of z for 97 percentage confidence interval is when we take a confidence interval of 97 percentage the critical value is value as 2 .17 the standard deviation we already know to be equal to 1 .3 minutes and what we don't know is the sample size right so so we are told in the question that this confidence interval should be must be within 0 .25 of the mean which means this means that the lower limit of this conference interval will be a difference of 0 .25 from the mean and the upper limit would also be 2 .0 .25 away from the mean.
02:27
So what we have what we have to find over here is that the sample size.
02:32
So we already know this part of the formula should be equal to 0 .25, right? so the critical value of z times the standard deviation divided by the square root of n is equal to 0 .25.
02:48
We know the critical value of z as to 0 .17.
02:52
The standard deviation is 1 .3 and what we have to find is n.
02:58
So we can rearrange this equation and what we will have is 2 .17 times 1 .3 divided by 0 .25 is equal to the square root of n.
03:12
So when we take 2 .17 and multiply it with 1 .3 and then divide it with 0 .25, we get the square root of n to be equal to 11 .284.
03:27
So, now we take the square of this number, so we take square on both sides and we get the value of n to be equal to a hundred and twenty seven point three.
03:39
So we approximate the sample size over here as one hundred and twenty seven.
03:45
So this is the first question that we needed to find.
03:49
So we have found the sample, the size of the sample over here.
03:54
Now let's move on to the second question, which is question number five.
03:57
Now in this question we are told that a random sample of 48 days in a recent year is taken and the us gasoline prices had a mean of 2 .34 dollars and a standard deviation of 0 .3 dollars.
04:16
So the random sample had 48 days and the mean that was calculated was a 2 .34 dollars and the random sample had 48 days and the mean that was calculated was 2 .34 dollars.
04:27
And the standard deviation is calculated to be $0 .3.
04:34
Now we are told to use this information to construct the following conference interval and we should also determine the margin of error.
04:45
So first we can write the formula 4 conference interval which is the mean x bar plus or minus the critical value of z times the standard deviation divided by the square root of n.
05:03
And what we call the margin of error in this formula is the second part.
05:09
That is, we call z the critical value of z times standard deviation divided by the square root of n as the margin of error.
05:21
So what we can see is that what we have told already in the first problem is that the confidence interval.
05:28
The z value for the confidence interval changes with what percentage of confidence interval we are taking.
05:35
So the first case over here is the confidence interval of 90 percentage.
05:43
So for 90 percentage the critical value of z is equal to of z is 1 .64.
05:55
So what we would have is we already have the mean of this data to be $2 .34.
06:08
We know that the critical z value is 1 .64.
06:12
We know the standard deviation is $0 .3 and the sample size is 48.
06:19
So when we calculate this, what we get over here as the confidence interval is, from 2 point from 2 .27 to 2 .27 to 2 .2 .41.
06:43
So this is what we get as the 90 percentage confidence interval.
06:49
Now the second part of the same question is to find the confidence interval for 95 percentage.
06:58
So what would change over here is just the critical value of z which would now be equal to for 95 percentage the value is 1 .96.
07:12
So the formula would be 2 .34 plus or minus 1 .96 times 0 .93 divided by the square root of 48 and on evaluating this we get the confidence interval to be equal to 2 .25 from 2 .25 to the upper limit is 2 .2 .42.
07:48
So this is what we get as the confidence 95 percentage confidence interval.
07:55
Now moving on to the third part of this question we are told to find the 99 percentage confidence interval as well.
08:04
So when the confidence interval we have to find this 99 percentage, the corresponding z value is 2 .58.
08:17
So now we can write the confidence interval is then equal to 2 .34 plus or minus 2 .58 times 0 .3 divided by the square root of 48.
08:32
So on evaluating this we get the required confidence interval as 2 .23 from 2 .23 to the upper limit over here is 2 .45 so this is the 99 percentage confidence interval.
08:54
So this is the 99 percentage confidence interval.
08:58
So that is all about the second question that is given to us and now let's move on to the third question which is the sixth question and it is similar to the fourth fifth question over here...