Answer:
The margin of error assuming a 90% confidence level is 0.3
Step-by-step explanation:
Size of the sample: n=320
Mean: m=24
Standard deviation: s=2.9
Confidence interval: 90%
100(1-α)=90
Solving for α: Dividing by 100 both sides of the equation above:
100(1-α)/100=90/100
1-α=0.9
Subtracting 1 both sides of the equation:
1-α-1=0.9-1
-α=-0.1
Multiplying the equation by -1:
(-1)(-α=-0.1)
α=0.1
n= [z(1-α/2) s / E]^2
where E is the margin of error
z(1-α/2)=z(1-0.1/2)=z(1-0.05)=z(0.95)=1.64 (Table standard normal distribution)
z(1-α/2)=1.64
Replacing the known values in the equation above:
320= [(1.64) (2.9) / E]^2
320= (4.756/ E)^2
Solving for E: Square root both sides of the equation:
sqrt(320)=sqrt[ (4.756/ E)^2]
17.88854382=4.756/E
Cross multiplication:
17.88854382 E = 4.756
Dividing both sides of the equation by 17.88854382:
17.88854382 E / 17.88854382 = 4.756 / 17.88854382
E=0.265868486
Rounding tho the nearest tenth:
E=0.3
