Problem 12 - Basic Statistics for Business and Economics with Student

Question

The research department at the home office of New Hampshire Insurance conducts ongoing research on the causes of automobile accidents, the characteristics of the drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident. At the .05 significance level, is there a significant difference in the proportions of single and married persons having an accident during a three-year period? Determine the \(p\) -value.

Step-by-Step Solution

Answer

No significant difference at 0.05 level; p-value is 0.104.

Step 1: Define the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis \( H_0 \) states that there is no difference in proportions of accidents between single and married persons, \( p_1 = p_2 \). The alternative hypothesis \( H_1 \) states that there is a difference, \( p_1 eq p_2 \).

Step 2: Calculate the Sample Proportions

Calculate the sample proportions for both groups. For singles, the proportion is \( \hat{p}_1 = \frac{120}{400} = 0.3 \). For married people, the proportion is \( \hat{p}_2 = \frac{150}{600} = 0.25 \).

Step 3: Compute the Pooled Proportion

Calculate the pooled proportion \( \hat{p} \) as follows: \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{120 + 150}{400 + 600} = \frac{270}{1000} = 0.27 \).

Step 4: Calculate the Standard Error

Calculate the standard error (SE) of the difference in proportions using the formula: \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.27 \times 0.73 \times \left(\frac{1}{400} + \frac{1}{600}\right)} \approx 0.0308 \).

Step 5: Compute the Test Statistic

Calculate the test statistic using the formula: \( Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.3 - 0.25}{0.0308} \approx 1.623 \).

Step 6: Determine the Critical Value and p-value

For a two-tailed test at \( \alpha = 0.05 \), the critical Z-value is \( \pm1.96 \). Calculate the p-value using the standard normal distribution. The p-value = 0.104, which is greater than \( \alpha = 0.05 \).

Step 7: Make a Decision

Since the p-value (0.104) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to suggest a significant difference in the proportions of accidents between single and married persons.

Key Concepts

Proportions

When we talk about proportions in statistics, we are discussing a type of ratio or fraction that describes a part of a whole. In the context of hypothesis testing, especially when comparing two groups, we focus on the sample proportions. For example, in the given problem, we want to understand if the proportion of single people having accidents differs from that of married people.
Let's break down how to calculate sample proportions:

• For singles: We found that 120 out of 400 had accidents. So, the proportion is calculated as \( \hat{p}_1 = \frac{120}{400} = 0.3 \).
• For married people: 150 out of 600 had accidents, giving a proportion of \( \hat{p}_2 = \frac{150}{600} = 0.25 \).

These proportions help us set up our hypothesis test to analyze if these differences are statistically significant. Keeping clear track of these proportions is crucial to understanding results from hypothesis testing, as they directly inform what we are measuring against.

P-value

The p-value represents the probability of finding a test statistic, like the Z-value in this example, as extreme as what was actually observed, assuming that the null hypothesis is true.
In simpler terms, it tells us how likely it is to see the observed data if our null hypothesis holds. Here’s how it plays into our example:

• After computing our test statistic, we ended up with a Z-value of approximately 1.623.
• We then determine the p-value from this Z-value, which in this case came out to be 0.104.
• Recall that for a predetermined significance level \( \alpha = 0.05 \), if the p-value is less than this significance level, we would reject the null hypothesis.

However, since our p-value is 0.104 (which is greater than 0.05), we conclude that there is insufficient evidence to suggest a significant difference. The beauty of the p-value is that it provides a quantitative measure to aid in decision-making by comparing it directly to your chosen significance level.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), represents a statement of no effect or no difference that we test against the data. In hypothesis testing for proportions, the null hypothesis often states that there is no difference between the two proportions involved. In this context:

• The null hypothesis \(H_0\) is that there is no difference in the proportion of single people and married people having accidents, which mathematically is expressed as \( p_1 = p_2 \).
• The alternative hypothesis, on the other hand, posits a difference, that is \( p_1 eq p_2 \).

The testing process involves gathering data and calculating test statistics to see if we have enough evidence to reject the null hypothesis. If the null hypothesis is rejected, it suggests that there is a statistically significant difference between the two proportions. In our example, since we didn’t have sufficient evidence (the p-value was above the significance level), we couldn’t reject the null hypothesis. Thus, we inferred that any apparent difference might have occurred due to random chance, not because there is an inherent true difference between the two groups.