A website reported that 33% of drivers 18 and older admitted to texting while dr

Question

A website reported that

33%

of drivers 18 and older admitted to texting while driving in 2009. In a random sample of

400

drivers 18 years and older drawn in 2010,

90

of the drivers said they texted while driving. Complete parts a through c below.

a. Construct a

99%

confidence interval to estimate the actual proportion of people who texted while driving in 2010.A

99%

confidence interval to estimate the actual proportion has a lower limit of

nothing

and an upper limit of

nothing.

(Round to three decimal places as needed.)

b. What is the margin of error for this sample?

The margin of error is

nothing.

(Round to three decimal places as needed.)

c. Is there any evidence that this proportion has changed since 2009 based on this sample?

Explanation / Answer

(a)

n = 400    

p = 90/400 = 0.225    

% = 99    

Standard Error, SE = {p(1 - p)/n} =    (0.225(1 - 0.225))/400 = 0.020879116

z- score = 2.575829304    

Width of the confidence interval = z * SE =     2.57582930354892 * 0.0208791163606126 = 0.05378104

Lower Limit of the confidence interval = P - width =     0.225 - 0.0537810397538735 = 0.17121896

Upper Limit of the confidence interval = P + width =     0.225 + 0.0537810397538735 = 0.27878104

The 99% confidence interval is [0.171, 0.279]

(b) Margin of error = 0.054

(c) Since the confidence interval excludes 0.33, there is sufficient evidence that the proportion has changed.

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