The National Highway Traffic Safety Administration (NHTSA) collects data on seat

Question

The National Highway Traffic Safety Administration (NHTSA) collects data on seat-belt use. Of 1,000 drivers 16-24 years old, 800 said they buckle up, whereas 933 of 1,120 drivers 25-50 years old said that they did. Do the data provide sufficient evidence to conclude that there is a significant difference in seat-belt use between the two age groups?
(a) State the research question.
(b) State both the null and alternative hypotheses.
(c) Determine the critical values (use 5% significance level).
(d) Calculate the standard error
(d) Calculate the test statistic, and state the decision regarding the null hypothesis.
(e) Calculate the p-value for the test.
(f) Assess the strength of your decision in part (d) using the p-value.
(g) Answer the research question. Which age group is more or less likely to use seat-belts, and by how much?
(h) Calculate a 95% confidence interval for the difference between the two corresponding population proportions
(i) How does your result in part (h) relate to the decision regarding the null hypothesis in parts (d), (e), and (f)? Be specific.

Explanation / Answer

Solution:-

p1 = 800/1000

p1 = 0.8

p2 = 933/1120

p2 = 0.833

a) There is sufficient evidence to conclude that there is a significant difference in seat-belt use between the two age groups.

b) 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. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

c) Critical values = + 1.96

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 = 0.8175

d)

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.01681

z = (p1 - p2) / SE

d) z = - 1.965

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 -1.965 or greater than 1.965

e) Thus, the P-value = 0.494

f) Interpret results. Since the P-value (0.494) is less than the significance level (0.05), we cannot accept the null hypothesis.

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