91影视

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Step by step solution

01

Identify Null and Alternative Hypotheses

The null hypothesis (H鈧�) and the alternative hypothesis (H鈧�) are appropriately stated for comparing the proportions, but ensure that their definitions align with the scenario described. The hypotheses should be: \[ H_0: p_1 - p_2 = 0 \] (meaning the pass rates are equal) and \[ H_a: p_1 - p_2 > 0 \] (meaning Instructor A has a higher pass rate).

02

Verify Proportion Calculations

The statement incorrectly calculates \(\hat{p}_1\) and \(\hat{p}_2\). The correct calculations are \(\hat{p}_1 = \frac{30}{50} = 0.60\) and \(\hat{p}_2 = \frac{22}{50} = 0.44\). Hence, double-check these values as they impact the pooled proportion and test statistic.

03

Check Pooled Proportion

Recalculate the pooled proportion, \(\hat{p}_C\), using the corrected individual success proportions. Therefore, \[ \hat{p}_C = \frac{30 + 22}{50 + 50} = 0.52. \] This calculation is actually accurate, so \(\hat{p}_C\) is correctly identified.

04

Test Statistic Calculation Verification

Recalculate the test statistic with the correct proportions: \( z = \frac{(0.60 - 0.44) - 0}{\sqrt{\frac{0.52(0.48)}{50} + \frac{0.52(0.48)}{50}}} \). This needs to be done precisely to determine if the test statistic is accurately computed.

05

Reevaluate Conclusion Based on Corrected Test Statistic and P-value

With a corrected test statistic, recompute the P-value. Compare this to the significance level (\(\alpha = 0.05\)). If the P-value is less than \(\alpha\), reject \(H_0\). Otherwise, do not reject. The original inference was incorrect if based on wrong calculations.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Significance Testing

Significance testing is a vital concept in statistics used to determine whether the observed data deviates significantly from what we'd expect under a certain hypothesis. In the context of the driving school problem, the significance test helps us determine if there's enough statistical evidence to support the claim that one instructor is more effective than the other.
To perform a significance test, follow these steps:

• State the null and alternative hypotheses.
• Choose a significance level, often denoted by \(\alpha\), typically set at 0.05 for a 5% level.
• Calculate the test statistic, which shows how much the data diverges from the null hypothesis.
• Determine the P-value, which indicates the probability of observing data at least as extreme as the actual observed data, assuming the null hypothesis is true.
• Finally, compare the P-value to \alpha\ to decide whether to reject or not reject the null hypothesis.

Employing these steps appropriately ensures that conclusions drawn from the results are scientifically and statistically valid.

Two-sample Z Test

The two-sample Z test is specifically designed to compare the proportions of two independent groups. In our driving school scenario, we want to compare the effectiveness of two instructors, both having different pass rates among their students.
Unlike single-sample tests, here we emphasize that the groups don't interact or influence each other, which justifies their independence. This ensures that our results can be generalized beyond our specific sample groups.
To conduct a two-sample Z test for proportions:

• Ensure that the samples are independent and randomly assigned.
• Calculate the sample proportions \( \hat{p}_1 \) and \( \hat{p}_2 \) for each group.
• Compute the pooled proportion, which combines the success data from both groups, serving as a point of comparison.
• Calculate the Z test statistic using the formula: \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\frac{\hat{p}_C(1-\hat{p}_C)}{n_1} + \frac{\hat{p}_C(1-\hat{p}_C)}{n_2}}} \]
  Here, \( n_1 \) and \( n_2 \) are the sizes of the two groups.
• Use the Z value to find the P-value, helping us determine if the difference between groups is statistically significant.

When done correctly, the two-sample Z test offers insights into whether differences observed between groups are of statistical importance.

Null and Alternative Hypotheses

In hypothesis testing, articulating the null and alternative hypotheses is crucial for a clear understanding of what the analysis aims to prove or disprove. For the driving school exercise, these hypotheses were:

• Null Hypothesis (\( H_0 \)): \( p_1 - p_2 = 0 \) 鈥� Indicates that there is no difference between the pass rates of students taught by Instructor A and Instructor B.
• Alternative Hypothesis (\( H_a \)): \( p_1 - p_2 > 0 \) 鈥� Suggests that Instructor A has a higher pass rate than Instructor B's students.

The null hypothesis is a statement of no effect or no difference, serving as the default assumption. Conversely, the alternative hypothesis posits a difference or an effect, directing the test's specific examination.
To test these, calculations (like the test statistic and P-values) revolve around these hypotheses. A statistical test's outcome will either support the null hypothesis by failing to provide sufficient evidence against it or reject it in favor of the alternative hypothesis when evidence is strong enough.

Pooled Proportion

The pooled proportion is a crucial part of the two-sample Z test, acting as an averaged proportion of success across the combined groups. In the driving school example, it's calculated by combining the successes and failures of both instructors.

• The formula for pooled proportion \( \hat{p}_C \) is: \[ \hat{p}_C = \frac{x_1 + x_2}{n_1 + n_2} \] \( x_1 \) and \( x_2 \) are the number of successes from each group, and \( n_1 \) and \( n_2 \) are the respective sizes of the groups.
• This pooled value helps to stabilize the test by providing a common proportion to base the variance upon, especially when sample sizes differ.
• It's essential since it aggregates the information from both groups, enabling us to compute the Z test statistic accurately.

In statistical tests, using a pooled proportion essentially allows for a more refined comparison by leveraging all available data, thus enhancing the test's reliability and robustness.