Question

The news release referenced in the previous exercise also included data from independent samples of teenage 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 news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the two populations. Do the data provide convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of \(0.05 .\)

Short answer

In order to provide a short answer, apply the aforementioned steps to the specific values given in the exercise, and then compare the result to the critical value to make a decision on whether to reject or fail to reject the null hypothesis.

Step-by-step solution

State the Hypotheses

The null hypothesis (H₀) is that the proportion of teens who approve (P₁) is the same as the proportion of parents who approve (P₂), i.e., P₁ - P₂ = 0. The alternative hypothesis (H₁) is that the proportion of teens who approve is less than the proportion of parents who approve, i.e., P₁ - P₂ < 0.

Calculate the Test Statistic

Firstly, find the overall sample proportion (p): \(p = (x₁ + x₂) / (n₁ + n₂)\), where x₁ is the number of successes in the first sample (teens who approve), x₂ is the number of successes in the second sample (parents of teens who approve), n₁ is the size of the first sample, and n₂ is the size of the second sample. Then, calculate the test statistic (Z) using the formula: \(Z = (P₁ - P₂) / √p(1 - p)(1/n₁ + 1/n₂) \), where P₁ and P₂ are the proportions of successes in the first and second sample.

Determine the Critical Value

For a significance level (α) of 0.05 and a one-tailed test, the critical value (Z₀) is -1.645. This value is obtained from a standard normal distribution table.

Make a Decision

Compare the test statistic (Z) with the critical value (Z₀). If Z < Z₀, then reject the null hypothesis. If Z ≥ Z₀, then fail to reject the null hypothesis.

Key concepts

Null and Alternative Hypotheses

In any study involving comparative analysis, formulating hypotheses is a foundational step. Hypotheses lay the groundwork on which statistical tests are performed. The null hypothesis, denoted as H₀, is a statement asserting there is no effect or no difference between the groups being studied. It serves as the initial assumption that any observed differences are due to random chance rather than an actual effect.

In contrast, the alternative hypothesis, denoted as H₁ or Hₐ, is the statement researcher is seeking evidence in favor of. It suggests that there is an effect or a difference, and the observed data are not solely due to chance. In the context of the given exercise, H₀ posits that the approval rates among teens and parents of teens are equal regarding the ban on cell phone and texting while driving. The alternative hypothesis contends that the approval rate among teens is lower than that of parents of teens. The rejection of the null hypothesis in favor of the alternative signifies that the data collected provides convincing evidence of a real difference.

Proportion Hypothesis Test

The proportion hypothesis test is a statistical technique used to infer whether there is a significant difference between the proportions of two groups. This type of test is particularly useful when the variable of interest is categorical, and we're concerned with the proportion or percentage outcomes falling into one of the categories.

In the case of the textbook problem, we're examining whether a significant difference exists between the proportions of teens and their parents who approve of laws banning cell phone use while driving. We calculate a test statistic to see if the observed proportion of teens who approve of the ban significantly deviates from that of their parents. This process involves assuming the null hypothesis is true and then determining the probability of observing a result as extreme as, or more extreme than, the one obtained if the null were indeed true. If this probability is lower than a predetermined significance level, we may have enough evidence to favor the alternative hypothesis, suggesting there is a significant difference.

Significance Level

The significance level, typically represented by the symbol α (alpha), is a threshold that determines how strong the evidence must be in order to reject the null hypothesis in a statistical test. It's an expression of the probability of making a Type I error, which is the mistake of rejecting a true null hypothesis. A common choice for α is 0.05, implying a 5% risk of concluding that a difference exists when there is none.

In the provided exercise, a significance level of 0.05 is used. This means that if the test statistic falls within the 5% extreme part of the distribution assuming the null hypothesis is true, the null hypothesis will be rejected. It correlates to the expected proportion of times that researchers would incorrectly reject the null hypothesis if they repeated the experiment an infinite number of times under the same exact conditions.

Test Statistic Calculation

The test statistic calculation is a crucial step in hypothesis testing. The test statistic, often denoted as Z, T, or F, depending on the test, is a standardized value used to decide whether to reject the null hypothesis. It measures how far the observed data falls from the null hypothesis, with a higher absolute value indicating a larger divergence.

To calculate the test statistic for a proportion hypothesis test, we first determine the pooled sample proportion from all groups combined. This proportion reflects the best estimate of the true proportion under the null hypothesis. Then, we find the difference between the two group proportions and standardize it using the standard error, which reflects the expected variability of the proportion differences due purely to sampling error. In the exercise, the test statistic is derived for a difference in proportions between teen and parent approval rates. The resultant Z-value will be compared against a critical Z-value from a standard normal distribution; if the calculated Z is more extreme than the critical value (in the direction predicted by the alternative hypothesis), the null hypothesis is rejected.