(a) The critical value for α = 0.1 \alpha=0.1 α = 0.1 is z c = z 1 − α / 2 = 1.6449. z_c=z_{1-\alpha/2}=1.6449. z c = z 1 − α /2 = 1.6449.
z c × σ n ≤ 1 z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1 z c × n σ ≤ 1
n ≥ ( z c σ 1 ) 2 n\geq(\dfrac{z_c\sigma}{1})^2 n ≥ ( 1 z c σ ) 2
n ≥ ( 1.6449 ( 12 ) 1 ) 2 n\geq(\dfrac{1.6449(12)}{1})^2 n ≥ ( 1 1.6449 ( 12 ) ) 2
n ≥ 390 n\geq390 n ≥ 390
(b) The critical value for α = 0.05 \alpha=0.05 α = 0.05 is z c = z 1 − α / 2 = 1.96. z_c=z_{1-\alpha/2}=1.96. z c = z 1 − α /2 = 1.96.
z c × σ n ≤ 1 z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1 z c × n σ ≤ 1
n ≥ ( z c σ 1 ) 2 n\geq(\dfrac{z_c\sigma}{1})^2 n ≥ ( 1 z c σ ) 2
n ≥ ( 1.96 ( 12 ) 1 ) 2 n\geq(\dfrac{1.96(12)}{1})^2 n ≥ ( 1 1.96 ( 12 ) ) 2
n ≥ 554 n\geq554 n ≥ 554
(c) The critical value for α = 0.1 \alpha=0.1 α = 0.1 is z c = z 1 − α / 2 = 1.6449. z_c=z_{1-\alpha/2}=1.6449. z c = z 1 − α /2 = 1.6449.
z c × σ n ≤ 1 / 2 z_c\times\dfrac{\sigma}{\sqrt{n}}\leq1/2 z c × n σ ≤ 1/2
n ≥ ( z c σ 1 / 2 ) 2 n\geq(\dfrac{z_c\sigma}{1/2})^2 n ≥ ( 1/2 z c σ ) 2
n ≥ ( 1.6449 ( 12 ) 1 / 2 ) 2 n\geq(\dfrac{1.6449(12)}{1/2})^2 n ≥ ( 1/2 1.6449 ( 12 ) ) 2
n ≥ 1559 n\geq1559 n ≥ 1559 The required sample size should be increased by 4 times to establish the new average life span to within 1/2 hours, with 90% level of confidence.
Comments