[Q13: Only answer if you ABSOLUTELY know how to solve AND the answers match up (

Question

[Q13: Only answer if you ABSOLUTELY know how to solve AND the answers match up ((tired of getting wrong replies.) So, please make sure you’re really certain. It’s appreciated.)

Really Explain EACH STEP, including any formulas you used (& how to use the formula), and explain how to compute with a TI-84 preferably/when possible.]

The Department of Transportion (DOT) is attempting to determine the proportion of drivers who require all passengers in the car to wear their seatbelt before putting the vehicle in drive. A survey of 84 drivers is performed and 30 people say they will not drive until all passengers in the vehicle are buckled up. To report their finding they want to create a 90% confidence interval. What would be the margin error for this confidence interval?

Question 13 options:

1)

0.0669

2)

0.0094

3)

0.0857

4)

0.0523

5)

0.0045

1)

0.0669

2)

0.0094

3)

0.0857

4)

0.0523

5)

0.0045

Explanation / Answer

p = 30/84 = 0.3571

n = 84

Z for 90% confidence interval = Z0.95 = 1.65

Margin of error = Z0.95 * sqrt (p*(1-p)/n)

= 1.65 * sqrt (0.3571 * (1-0.3571)/84)

= 0.0857

Option-3) 0.0857

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