Question

suppose that an investigator wishes to estimate the content of mercury in the water of a certain area. He collected a measurements and obtained the following data:
409, 406,400,399,402,406,401,403,398.

construct 95℅ confidence interval of mean content mercury in the water of a certain area.​

Answer 1

Answer:

Let the same common difference of two AP’s isd, Given that, the first term of first AP and second AP are 2 and 7 respectively, then the AP’s are

2,2 + d,2 + 2d,2 + 3d,.,.

and 7,7+ d, 7 +2d, 7+3d,…

Now, 10th terms of first and second AP’s are 2 + 9d and 7 + 9 d, respectively.

So, their difference is 7 + 9d – (2 + 9d) = 5

Also, 21st terms of first and second AP’s are 2 + 20d and 7 + 20d, respectively.

So, their difference is 7 + 20d – (2 + 9d) = 5

Also, if the a„ and bn are the nth terms of first and second AP.

Then, bn -an = [7 + (n-1)d)] – [2 + (n-1)d] = 5

Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.

Answer 2

Explanation:

suppose that an investigator wishes to estimate the content of mercury in the water of a certain area. He collected a measurements and obtained the following data:

409, 406,400,399,402,406,401,403,398.

construct 95℅ confidence interval of mean content mercury in the water of a certain area.