Among 320 randomly selected airline travelers, the mean number of hours spent travelling per year is 24 hours and the standard deviation is 2.9. what is the margin of error, assuming a 90% confidence level? round your answer to the nearest tenth. 0.01

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

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step-by-step explanation:

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