After paying $5.12 for a salad, Norachai has $27.10. How much money did he have
2 answers:
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Answer:
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
