Question

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation of all components is 2.2 mm, how many of these components should she consider to be 80% sure of know the mean will be within 0.3 mm? Group of answer choices 1080 126 10 80 Way more than any of these answers

Answer 1

Answer:

The correct option is;

80

Step-by-step explanation:

The standard deviation of the components = 2.2 mm

The difference in the mean = 0.3

The level of confidence (power)= 80%

The formula for finding the sample size is given as follows;

$n = \dfrac{2 \times \left [ (a + b)^2 \right ] \times \sigma ^2}{\left (\mu_1 - \mu_2\right )^2}$

Where;

μ₁ - μ₂ = Is the difference in the mean = 0.3

a = The α multiplier = 0.05

b = The power multiplier = 0.8

σ = The standard deviation

n = The sample size

By substituting in the values, we have;

$n = \dfrac{2 \times \left [ (0.05 + 0.8)^2 \right ] \times 2.2 ^2}{\left (2.2\right )^2} = 77.7$

n ≈ 8

Rounding up to the next 10th gives;

n = 80

Therefore, the correct sample size should be about 80