Question

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A recent random survey of 100 individuals in Michigan found that 80 drove to work alone. A similar survey of 120 commuters in New York found that 62 drivers drove alone to work. Find the \(95 \%\) confidence interval for the difference in proportions.

Short answer

There is a significant difference in the proportions of people driving alone to work between Michigan and New York.

Step-by-step solution

State the Hypotheses

The null hypothesis \( H_0: p_1 - p_2 = 0 \) states that there is no difference in the proportions of individuals in Michigan who drive alone to work compared to those in New York. The alternative hypothesis \( H_a: p_1 - p_2 eq 0 \) suggests that there is a difference in these proportions.

Identify the Claim

The claim is that there is a difference in the proportions of people who drive to work alone between Michigan and New York, as denoted by the alternative hypothesis \( H_a \).

Find the Critical Value

For a two-tailed test at the \(95\%\) confidence level, the critical value in the standard normal distribution is \( z = 1.96 \). This means any test statistic beyond \( \pm 1.96 \) will lead us to reject the null hypothesis.

Compute the Test Value

First, calculate the sample proportions:\( \hat{p_1} = \frac{80}{100} = 0.8 \) (Michigan)\( \hat{p_2} = \frac{62}{120} \approx 0.5167 \) (New York).The pooled sample proportion \( \hat{p} \) is:\[ \hat{p} = \frac{80 + 62}{100 + 120} = \frac{142}{220} \approx 0.6455 \]The standard error \( SE \) of the sampling distribution is:\[ SE = \sqrt{ \hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right) } \]\[ = \sqrt{ 0.6455(1 - 0.6455) \left(\frac{1}{100} + \frac{1}{120}\right) } \approx 0.0722 \]The test statistic \( z \) is:\[ z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.8 - 0.5167}{0.0722} \approx 3.92 \]

Make the Decision

Since the test statistic \( z \approx 3.92 \) is greater than the critical value \(1.96\), we reject the null hypothesis \( H_0 \). This indicates that there is a statistically significant difference between the proportions of individuals driving to work alone in Michigan and New York.

Summarize the Results

The results suggest that it is statistically significant to claim a difference in the percentage of people driving alone to work between Michigan and New York at the \(95\%\) confidence level.

Key concepts

Confidence Interval

A confidence interval provides a range of values within which we can be fairly certain the true population parameter lies. In this exercise, we're looking at the difference in proportions of people who drive alone to work between Michigan and New York at a 95% confidence level.
The key here is the term "95% confidence level." It means that if we were to take 100 different samples and construct confidence intervals for each one, approximately 95 of those intervals would contain the true difference in proportions.
To find the confidence interval, you calculate the difference between the two sample proportions (20% for Michigan and approximately 51.67% for New York). Against this observed difference, you apply the critical value and standard error to construct the interval bounds. Understanding this allows us to determine whether the observed difference is statistically significant.

Critical Value

The critical value is a key concept in hypothesis testing that defines the threshold at which you will decide whether to reject the null hypothesis. In a two-tailed test such as this, determining the critical value involves looking at the standard normal distribution.
At a 95% confidence level, the critical value for a two-tailed test is 1.96. This means if your test statistic is greater than 1.96 or less than -1.96, you will reject the null hypothesis. The 1.96 value is derived from the z-distribution which indicates that outcomes outside of this value occur with less than 5% probability under the null hypothesis. This helps ensure that we are making conclusions with high confidence.

Null Hypothesis

The null hypothesis, denoted by \( H_0 \), is a statement that there is no effect or no difference. It serves as the starting assumption in hypothesis testing.
In this exercise, the null hypothesis is \( H_0: p_1 - p_2 = 0 \), which means we initially assume there is no difference in the proportion of people driving to work alone in Michigan versus New York.
The purpose of the null hypothesis is to provide a specific claim that can be tested. By either rejecting or failing to reject \( H_0 \), one can determine if there is enough statistical evidence to support an alternative hypothesis \( H_a \), which is the claim of a difference in this case.

Test Statistic

A test statistic is a standardized value that you calculate from sample data during a hypothesis test. It is used to compare your data to what is expected under the null hypothesis.
In this situation, the test statistic is calculated using the formula: \[ z = \frac{\hat{p_1} - \hat{p_2}}{SE} \]where \(\hat{p_1}\) and \(\hat{p_2}\) are the sample proportions for Michigan and New York, and \(SE\) is the standard error of the difference in proportions.
In our context, the calculated test statistic is approximately 3.92. This means that the observed difference between sample proportions is 3.92 standard deviations away from no difference (the null hypothesis value), which is highly significant considering it exceeds the critical value of 1.96.