The sample size such that the sample mean does not differ by more than 2 units is 21.12
confidence level = 95 percent = 0.95
Let the sample size be n
where z_α/2 is the critical value at α/2
standard deviation = σ
we know standard deviation = √variance = √22 = 4.69
The margin of error E = 2
for 95% confidence level α = 1 - 0.95 = 0.05
z_(0.05/2) = z_ 0.025 = 1.96
μ is the mean of the test scores
We need to find the sample size n such that probability
P(| x - μ| < 2) = 0.95
⇒P(-2 < x - μ < 2) = 0.95
⇒P( -2/σ/√n < x -μ / (σ/√n) < 2/σ/√n) = 0.95
⇒P( -2√n / σ < z < 2√n / σ) = 0.95 (because z = x- μ / (σ/√n) ) (1)
Consider
P(- z_α/2 < z < z_α/2) = 0.95
⇒P( -1.96 < z < 1.96) =0.95 (2)
From (1) and (2)
2√n / σ = 1.96
⇒n = (1.96 * σ / 2)² = (1.96 * 4.69 / 2 )² = 21.12
Thus the sample size n = 21.12
A problem based on confidence level
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