Question

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.

Answer 1

Answer:

.3

Greater than .3

Step-by-step explanation:

Edge 100%

Answer 2

The margin of error of the random selection is 0.29

The given parameters are:

$n = 320$ --- the sample size

$\sigma = 2.9$ --- the standard deviation

$\bar x = 24$ --- the mean

$\alpha = 90\%$ --- the confidence level.

The margin of error (E) is calculated as follows:

$E = z \times \sqrt{\frac{\sigma^2}{n}}$

So, we have:

$E = z \times \sqrt{\frac{3.2^2}{320}}$

$E = z \times \sqrt{\frac{10.24}{320}}$

The z-value for 90% confidence level is 1.645.

Substitute 1.645 for z

$E = 1.645 \times \sqrt{\frac{10.24}{320}}$

$E = 1.645 \times \sqrt{0.032}$

Take square roots

$E = 1.645 \times 0.1789$

Multiply

$E = 0.2943$

Approximate

$E = 0.29$

Hence, the margin of error is 0.29

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