Question

part b: what is the margin of error for this sample ?

part c: is there any evidence that this proportion has changed since 2009 based on this sample ?

A website reported that 32% of drivers 18 and older admitted to texting while driving in 2009. In a random sample of 200 drivers 18 years and older drawn in 2010, 39 of the drivers said they texted while driving Complete parts a through c below a. Construct a 95% confidence interval to estimate the actual proportion of people who texted while driving in 2010 A95% confidence interval to estimate the actual proportion has a lower limit of and an upper limit of (Round to three decimal places as needed.)

Answer 1

Answer)

Po = 0.32

N = 200

First we need to check the conditions of normality that is if n*po and n*(1-po) both are greater than 5 or not

N*p = 64

N*(1-p) = 136

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 95% confidence level is 1.96

Margin of error (MOE) = Z*standard error

Standard error = √{po*(1-po)}/√n

Po = 0.32

N = 200

Z = 1.96

After substitution

MOE = 0.06465029620

Confidence interval is given by

Lower limit = P - MOE

P = 39/200

LOWER LIMIT = 0.13034970379 ~ 0.13

Upper limit = P + MOE = 0.25965029620 = 0.26

B)

Margin of error = 0.06465

C)

Null hypothesis Ho : P = 0.32

Alternate hypothesis Ha : P not equal to 0.32

As the interval do not contain the value 0.32

We reject Ho

So we have enough evidence to support the claim that percentage is changed