00:01
Hello student, as per the given problem first we will note down the given info.
00:05
So we are provided with a total sample size n to be 100, number of chips from the operation a in the sample, number of chips from the operation b in the sample, number of defective chips in the operation a and b respective.
00:19
So first as per the question we have to solve the problem by simple random sample estimate.
00:25
So to estimate the proportion of defective in the lot using a simple random sample we can use the sample.
00:32
By using the given information first we will calculate the p hat total.
00:37
So formula for p hat total is equal to total defective chips in the sample divided by total sample.
01:07
Now we will plug in value here and it becomes total defective chips in the sample is equal to 6 plus 10 divided by 100.
01:16
So it comes out to be 0 .16.
01:18
So we can say that p hat total is 0 .16.
01:23
So the estimated proportion of defective in the lot is 0 .16.
01:28
Next bound on the error of error.
01:31
So to place the bound of error we can calculate the confidence interval for the proportion using the confidence level.
01:43
For that we can use the formula margin of error is equal to z into under root of p hat total multiplied by 1 minus p hat total divided by n.
02:00
Now z is the critical value which we can calculate by the desired confidence.
02:06
Here it is given to be 95 percent.
02:11
So our z come out to be 1 .96 into under root of 0 .16 into 0 .84 divided by 100...