Question

The press release referenced in the previous exercise also included data from independent surveys of teen- age drivers and parents of teenage drivers. In response to a question asking if they approved of laws banning the use of cell phones and texting while driving, 74% of the teens surveyed and 95% of the parents surveyed said they approved. The sample sizes were not given in the press Bold exercises answered in back Data set available online release, but for purposes of this exercise, suppose that 600 teens and 400 parents of teens responded to the surveys and that it is reasonable to regard these samples as representative of the two populations. Do the data provide convincing evidence that the proportion of teens that approve of cell-phone and texting bans while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of .05.

Answer 1

It is given that the study of independent surveys of teen age drivers and parents of teenage drivers. 74% of the teens surveyed and 95% of the parents surveyed said they approved, response given for a question asking if they approved of laws banning the use of cell phones and texting while driving Suppose 600 teens and 400 parents of teens responded to the surveys and that it is reasonable to regard these samples as representative of the two populations. We have to test whether the data provide any sufficient evidence to conclude that the proportion of teens that approve of cell-phone and texting bans while driving is less than the proportion of parents of teens who approve, at the level of significance a 0.05 The proportion of teens that approve of cell-phone and texting bans while driving is P = 0.74 (or 74%) The proportion of parents of teens who approve of cell-phone and texting bans while driving is p 0.95 (or 95%) The combined estimate of p, is given below пр+п,р. 600(0.74)+400(0.95) 600400 = 0.82

Step1: p, Proportion of teens that approve of cell-phone and texting bans while driving p, Proportion of parents of teens who approve of cell-phone and texting bans while driving Step2: The null hypothesis The data provide no sufficient evidence to conclude that the proportion of teens that H approve of cell-phone and texting bans while driving is less than the proportion of parents of teens who approve, at the level of significance a = 0.05 That is, H p - P2 = 0 Step3 The alternative hypothesis H The data provide sufficient evidence to conclude that the proportion of teens that approve of cell-phone and texting bans while driving is less than the proportion of parents of teens who approve, at the level of significance a 0.05 That is, H p-P2< 0 Step4: Significance level is a 0.05

Step5: The formula of the test for two population proportions is given below. Test statistic, z(P\- P2)-hypothesis value P.(1-p)(1-.) + п, п, Step6: Assumptions: The two samples were independently selected. It is reasonable to regard the samples as representative of the two populations of interest. Checking to make sure that the sample sizes are large enough by using n600, P, = 0.74, n, =400 and p 0.95. we have n,P44410 n (1-156 > 10 п,р, 3 380 >10 п, (1-р,) %— 20 >10 Step7: Calculations: The analysis of two-proportion z-test is obtained, by using Minitab software

Calculations: The analysis of two-proportion z-test is obtained, by using Minitab software. Instructions: 1. Click Stat, Basic Statistics, and 2 Proportions... 2. Select Summarized data.... 3. Enter the values of Events, Trials for the First and Second Samples in their corresponding boxes. 4. Click on Options... 5. In the Confidence level box, type 0.95 and enter in Test Difference box, type the hypothesized two proportion difference (0), and specify the Alternative (less than) 6. Tick in the select box of use pooled estimate of p for test From the Minitab output, the two-proportion z-test shown below.

Test and CI for Two Proportions N Sample p Sample X 444 600 0.740000 2 380 400 0.950000 Difference p (1) -p (2) -0.21 Estimate for difference: 95 upper bound for difference Test for difference 0 (vs< 0): -0.175520 Z -8.54 P-Value = 0.000 The test statistic from the Minitab output is -8.54. Step8: The P-value for the two-proportion z-test at 5% level of significance is 0.043 That is, P(2-8.54) * 0.000 The decision rule: If P-value a then we reject our null hypothesis, otherwise we fail to reject our null hypothesis Ho The P-value for the given data is 0.000, which is less than the level of significance a 0.05 Therefore, we reject our null hypothesis H Therefore, the data provide sufficient evidence to conclude that the proportion of teens that approve of cell-phone and texting bans while driving is less than the proportion of parents of teens who approve, at the level of significance a 0.05