Question

A study was conducted to compare the proportion of drivers in Boston and New York who wore seat belts while driving. Data were collected, and the proportion wearing seat belts in Boston was 0.581 and the proportion wearing seat belts in New York was 0.832.

Due to local laws at the time the study was conducted, it was suspected that a smaller proportion of drivers wear seat belts in Boston than New York.

(a) Find the test statistic for this test using Ha: p_B < p_NY. (Use standard error = 0.085.)
(3 decimal places)
(b) Determine the p-value.
(3 decimal places)
(c) Based on this p-value, which of the following would you expect for a 95% confidence interval for p_B - p_NY?

All values in the interval are negative. Some values in the interval are positive and some are negative.    All values in the interval are positive.

(d) What is the correct conclusion?

( )The proportion of Boston drivers wearing seat belts was significantly lower than the proportion of New York drivers wearing seat belts.

( )The proportion of Boston drivers wearing seat belts was significantly higher than the proportion of New York drivers wearing seat belts.    

( )There was no significant difference between the proportion of drivers wearing seat belts in Boston and New York.

Answer 1

Given that the proportion wearing seat belts in Boston was 0.581 and the proportion wearing seat belts in New York was 0.832. During the study, it was suspected that a smaller proportion of drivers wear seat belts in Boston than New York. Let 3 be the proportion of people wearing seat belts in Boston, that is, z =0.581 and let Øvy be the proportion of people wearing seat belts in New York, that is, ÔNr =0.832. (a) The null and alternative hypotheses are given as follows: H: PBPNY H: P3 <P NY The standard error is given to be 0.085. The test statistic is given as follows: 7- PB - Ê NY SE(PB- ) 0.581-0.832 0.085 =-2.593 Thus, the test statistic is -2.593). (b)

The p-value for the left-tailed test can be computed in MS Excel as follows: fx -NORM.S.DIST(-2.953,TRUE) E F 0.001574) Thus, the p-value is 0.002). (c) Since the p-value is less than the level of significance (0.05), therefore, the correct option for 95% confidence interval for Papy is All values in the interval are negative. (d) The decision rule is to reject the null hypothesis if the p- value is less than the level of significance. Here the p-value (0.002) is less than the level of significance (0.05), therefore, we reject the null hypothesis and conclude that the proportion of Boston drivers wearing seat belts was significantly lower than the proportion of New York drivers wearing seat belts.