Question

A city wants to know if a new advertising campaign to make citizens aware of the...

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 Di 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 (?, ? ଶ ) distribution. Information collected randomly during a week before and a week before and a week a month after the campaign indicate a mean difference ?̅= −2 , with 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

b. You are asked by the city administration to study whether the advertising campaign was effective. State in terms of u , the relevant null and alternative hypothesis in conducting this study.

c. Compute the t statistic for testing H0 against HA

d. Obtain the p- value for the test

e.Do you reject H0 at the 5% level? At the 1% level?

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

Homework Answers

Answer #1

(a) The 95% confidence interval is:

= xd ± t*(sd/√n)

= -2± 2.45*(3.162/√7)

= -4.92, 0.92

The 95% confidence interval for the true average difference in number of drivers stopped with excess alcohol in their systems is between -4.92 and 0.92.

(b) The hypothesis being tested is:

H0: µd = 0

Ha: µd > 0

(c) The test statistic, t = xd/sd/√n

t = -2/3.162/√7

t = -1.67

(d) The p-value is 0.0726.

(e) Since the p-value (0.0726) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Since the p-value (0.0726) is greater than the significance level (0.01), we fail to reject the null hypothesis.

(f) Therefore, we cannot conclude that the advertising campaign was effective.

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