Refer previous problem. Suppose that you wish to estimate the difference between the mean acidity for rainfalls at two different locations. If you wish your estimate to be correct to the nearest 0.1 pH with probability near 0.90, approximately how many rainfalls (pH values) must you include in each sample?

Hence,we need at least 136 rainfall PH values in the sample i.e

n ≥ 136

Step-by-step explanation:

We are given that:

(σ1)^2 = (σ2)^2 = Population variance = 0.25

So, E < 0.1

Confidence coefficient (c) = 0.9

n = n1 = n2

For confidence level, 1 - α = 0.9,we'll determine Z (α /2) = Z 0.05 by looking up 0.005 using the normal probability table which i have attached.

So, Z (α /2) = 1.645

The margin of error E is given as;

E = Z (α /2)√[(σ1)^2)/n1] + [(σ2)^2)/n2]

= Z (α /2)√({(σ1)^2 + (σ2)^2}/n) < 0.1

Multiply both sides by √n to get;

Z (α /2)√(σ1)^2 + (σ2)^2} < 0.1√n

Divide both sides by 0.1;

{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1 <√n

When we square each side, we get

{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1} ^2 < n

We'll now fill in the known values and solve;

n > ( 1.645 x √{(0.25 + 0.25)/0.1}^2

n > 135.3 or approximately n > 136

Hence,we need at least 136 observations in the sample i.e

n ≥ 136

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.