Studies show that drivers who use cell phones while driving increase their risk of an accident. In a sample of 371 cases of new drivers using a cell phone while driving, 24 resulted in a crash. Find a 99% confidence interval estimate of the true population proportion of new drivers who were in a crash while using their cell phone.

The 99% confidence interval estimate of the true population proportion of new drivers who were in a crash while using their cell phone is 0.0647 +/- 0.0330 = (0.0317, 0.0977)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 24/371 = 0.0647

Number of samples n = 371

Confidence interval = 99%

z(at 99% confidence) = 2.58

Substituting the values we have;

0.0647 +/- 2.58√(0.0647(1-0.0647)/371)

0.0647 +/- 0.032950374780

0.0647 +/- 0.0330

(0.0317, 0.0977)

The 99% confidence interval estimate of the true population proportion of new drivers who were in a crash while using their cell phone is 0.0647 +/- 0.0330 = (0.0317, 0.0977)

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step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b