Among 320 randomly selected airline travelers, the mean number of hours spent travelling per year is 24 hours and the standard d

Question

2 years ago

8

eviation is 2. 9. What is the margin of error, assuming a 90% confidence level? Round your answer to the nearest tenth.

2 answers:

Andre45 [30] · Answer 1 · 2023-02-06T19:38:38+03:00

7 0

Answer:

.3

Greater than .3

Step-by-step explanation:

Edge 100%

Ostrovityanka [42] · Answer 2 · 2023-02-06T17:14:54+03:00

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