Question

The following data come from a study designed to investigate drinking problems among college students. In 1983, a group of students were asked whether they had ever driven an automobile while drinking. In 1987, after the legal drinking age was raised, a different group of college students were asked the same question. SHOW EXCEL CODES

Drove While Drinking Year

1983 1987 Total

Yes 1250 991 2241

No 1387 1666 3053

Total 2637 2657 5294

A. Use the chi-square test to evaluate the null hypothesis that population proportions of students who drove while drinking are the same in the two calendar years.

B. What do you conclude about the behavior of college students?

C. Again test the null hypothesis that the proportions of students who drove while drinking are identical for the two calendar years. This time, use the method based on the normal approximation to the binomial distribution that was presenting in Section 14.6. Do you reach the same conclusion?

D. Construct a 95% confidence interval for the true difference in population proportions.

E. Does the 95% confidence interval contain the value 0? Would you have expected it to?

Answer 1

Solution:-

A)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: Population proportions of students who drove while drinking are the same in the two calendar years

Alternative hypothesis: At least one of the null hypothesis statements is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for homogeneity.

Analyze sample data. Applying the chi-square test for homogeneity to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1)
D.F = 1
E_r,c = (n_r * n_c) / n

Χ² = 55.36

where DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, n_r is the number of observations from population r, n_c is the number of observations from level c of the categorical variable, n is the number of observations in the sample, E_r,c is the expected frequency count in population r for level c, and O_r,c is the observed frequency count in population r for level c.The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 55.36.

We use the Chi-Square Distribution Calculator to find P(Χ² > 55.36) = less than 0.0001.

Interpret results. Since the P-value (amost 0) is less than the significance level (0.05), we reject the null hypothesis.

B) From the above test we can conclude that population proportions of students who drove while drinking are different in the two calendar years.

C)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.42331
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.01362
z = (p₁ - p₂) / SE

z = 7.42

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -7.42 or greater than 7.42.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we can conclude that population proportions of students who drove while drinking are different in the two calendar years.

D) From the above test we can conclude that population proportions of students who drove while drinking are different in the two calendar years is C.I = ( 0.0744 , 0.1278).

C.I = (0.10105) + 1.96*0.01362

C.I = 0.10105 + 0.0267

C.I = ( 0.0744 , 0.1278)

E) No, 95% confidence interval does not contain the value 0, we also not expected it, because there is significance difference in the proportions.