Question

1. A survey of 1000 drivers found that 29% of the people send text messages while driving. Last year, a survey of 1000 drivers found that 17% of those sent text messages while driving.

(Show your work without using software to solve)

a. Give a 95% confidence interval for the increase in text messaging while driving. b. At α = .05, can we conclude that there has been an increase in the number of drivers who text while driving? State the hypotheses, find the test statistic, and state your conclusion and the reasoning behind it.

Answer 1

Part a

We are given

Population proportion = p = 0.17

Sample proportion = P = 0.29

Sample size = n = 1000

Confidence level = 95%

Confidence interval for Population Proportion

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Z = 1.96 (by using z-table)

Confidence interval = 0.29 ± 1.96*sqrt(0.29*(1 - 0.29)/1000)

Confidence interval = 0.29 ± 1.96* 0.014349

Confidence interval = 0.29 ± 0.0281

Lower limit = 0.29 - 0.0281 = 0.2619

Upper limit = 0.29 + 0.0281 = 0.3181

There is sufficient evidence to conclude that there has been an increase in the number of drivers who text while driving, because the value 0.17 is not lies within above interval.

Now, we have to perform z test for population proportion.

H₀: There has been no increase in the number of drivers who text while driving.

H_a: There has been an increase in the number of drivers who text while driving.

H₀: p = 0.17 vs. H_a: p > 0.17

Z = (P – p)/sqrt(p*(1 – p)/n)

Z = (0.29 – 0.17)/sqrt(0.17*(1 – 0.17)/1000)

Z = 10.1022

P-value = 0.00

(by using z-table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there has been an increase in the number of drivers who text while driving.