Question

II-(7 pts) A city wants to know if a new advertising campaign to make citizens aware of the dangers of

driving after drinking has been effective. They count the number of drivers who have been stopped with

more alcohol in their systems than the law allows for each day of the week in the week before and the

week a month after the campaign starts. Let ?? be the difference between the number of drivers caught

with excessive alcohol in their systems before and after the campaign on each calendar day in any given

week. Treat these as a random sample from a Normal (?, ??) distribution.

Information collected randomly during the seven days of a week before and a week a month after the

campaign, indicate a mean difference ?̅ = −2 , and a standard deviation ? = 3.162.

a. Obtain a 95% confidence interval for the true average difference in number of drivers stopped with

excess alcohol in their systems. (1pt)

b. You are asked by the city administration to study whether the advertising campaign has been

effective. State in terms of , the relevant null and alternative hypothesis in conducting this study. (1pt)

c. Compute the t statistic for testing ?? against ?? (1pt)

d. Obtain the p- value for the test (1pt)

e. Do you reject ?? at the 5% level? At the 1% level? (1pt)

f. Provide a short summary of your conclusions from this study. Comment on the practical versus

statistical significance of this estimate. (2pts)

w

1 w

b R b R

b R w

1 w s b R  wR  (1 w)R



Answer 1

Given data:

A city wants to know if a new advertising campaign to make citizens aware of the dangers of driving after drinking has been effective.

Let

? = be the difference between the number of drivers caught

Treat these as a random sample from a Normal (?, ??) distribution.

Mean difference ?̅ = −2

Standard deviation ? = 3.162.

a) Obtain a 95% confidence interval for the true average difference in number of drivers stopped with excess alcohol in their systems:

Sample mean difference (xbar) = -2

Sample Standard deviation ? = 3.162

Degrees of freedom = 7-1 = 6

            T for 95% confidence interval = 2.446899

            Confidence interval = xbar±ts/n

                                            = -2± 2.4469*3.162/7

            Confidence interval = (4.9244, 0.9243)

b) State in terms of , the relevant null and alternative hypothesis in conducting this study:

A hypothesis is a suggested solution for an unexplained occurrence that does not fit into current accepted scientific theory.

                        Null hypothesis: H_{0 :µd =} 0

                        Alternate hypothesis: H_a :_{µd >} 0

c) the t statistic for testing ?? against ??:

t = xbar- _µd/sn

= -2-0/3.162/7

t = -1.6735

d) Obtain the p- value for the test:

P value for t =-1.6735

Degree of freedom = 6 is 0.072979

e) Do you reject ?? at the 5% level? At the 1% level?

At 5% significance levels as a P value (0.072979)>0.65; therefore we can reject the Null Hypothesis.

At 1% significance level as a P value (0.072979)>0.01; therefore we can reject the Null Hypothesis.

f) In summary we can say that there is no efficient evidence to say that the number of drivers caught with the excessive alcohol has reduced.