Question

DO NOT HANDWARITE PLEASE

DO NOT HANDWARITE PLEASE

1. Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured. A different random sample of 1,100 drivers found 96.9% compliance.

1. State an appropriate null and alternative hypothesis for testing whether or not there is any statistical difference (i.e., a two-sided test) in these two proportions measured initially and then one year later. Conduct the test of hypothesis using a significance level of α= 0.05.USE R FOR THE TEST STATISTIC. Be sure to check the assumptions and conditions for your test. State the P-value of your test and also state your conclusion. (Feel free to use R in showing your work - and show your commands and output if using R.)
2. Develop a 95% confidence interval for the true difference in proportions between the first survey and the second survey and explain what this confidence interval means in context of this problem. (Feel free to use R in showing your work - and show your commands and output if using R.

Answer 1

Null Hypothesis H0: p1 = p2

Alternative Hypothesis H0: p1 p2

where p1 and p2 are the true proportions of people in compliance with law initially and after 1 year period.

For sample 1,

x1 = n1p1 = 1250 * 0.989 = 1236.25

n1(1-p1) = 1250 * (1 - 0.989) = 13.75

For sample 2,

x2 = n2p2 = 1100 * 0.969 = 1065.9

n1(1-p1) = 1100 * (1 - 0.969) = 34.1

Since, np > 10 and n(1-p) > 10, for both the samples ,  conditions for two proportion z test is satisfied.

Running two proportion z test in R, we get

n1 = 1250; n2 = 1100;
p1 = 0.989; p2 = 0.969
#Pooled sample proportion
p = (n1*p1+n2*p2)/(n1 + n2)
#standard error of sample proportion
se = sqrt(p*(1-p)*((1/n1)+(1/n2)))
#Test statistic,
z = (p1 - p2)/se
#p-value
p.value = 2 * pnorm(z, lower.tail = FALSE)

Test statistic,

> z
[1] 3.42537

P-value,

> p.value
[1] 0.0006139626

> se
[1] 0.005838785

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that p1 p2.

Margin of error = z * se = 1.96 * 0.005838785 = 0.01144402

95% confidence interval is,

Lower bound =  (p1 - p2) - margin of error = (0.989 - 0.969) - 0.01144402 = 0.00855598

Upper bound = (p1 - p2) + margin of error = (0.989 - 0.969) + 0.01144402 = 0.03144402

We're 95% confident that the interval (0.00855598, 0.03144402) captured the true value of p1 - p2.