/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 A state's Division of Motor Vehi... [FREE SOLUTION] | 91Ó°ÊÓ

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A state's Division of Motor Vehicles (DMV) claims that \(60 \%\) of teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the \(\alpha=\) 0.05 significance level that the DMV's claim is incorrect?

Short Answer

Expert verified
Yes, the evidence suggests the DMV's claim is incorrect at the 0.05 significance level.

Step by step solution

01

State the Hypotheses

We are conducting a hypothesis test to determine if the DMV's claim that 60% of teens pass their driving test on the first attempt is incorrect. Let \( p \) be the actual proportion of teens who pass on their first attempt. The null hypothesis \( H_0: p = 0.60 \) states that the true proportion is 60%, while the alternative hypothesis \( H_a: p eq 0.60 \) suggests that the proportion is different from 60%.
02

Calculate the Test Statistic

The test statistic is calculated using the formula for a proportion: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}}\]where \(\hat{p} = \frac{86}{125} = 0.688\), \( p_0 = 0.60 \), and \( n = 125 \). Plugging in these values, we get:\[z = \frac{0.688 - 0.60}{\sqrt{\frac{0.60 \times 0.40}{125}}} = 2.575\]
03

Determine the Critical Value and P-value

For a two-tailed test with \(\alpha = 0.05\), we look at the standard normal distribution to find the critical value, which is \(z^* \approx ±1.96\). Next, we find the p-value by looking up the z-score of 2.575 in the standard normal distribution table, which gives us a p-value of approximately 0.010. Since this is a two-tailed test, the p-value needs to be doubled: \(2 \times 0.010 = 0.020\).
04

Make a Decision

Compare the p-value with the significance level \(\alpha = 0.05\). Since \(0.020 < 0.05\), we reject the null hypothesis \(H_0: p = 0.60\). This indicates that there is significant statistical evidence to suggest that the DMV's claim is incorrect.

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Key Concepts

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

Significance Level
The significance level is a crucial component in hypothesis testing. It is denoted by the Greek letter Alpha (\(\alpha\)). The significance level represents the probability of rejecting the null hypothesis when it is true. Essentially, it is the threshold we set for determining whether a test result is statistically significant or not.
The lower the significance level, the stricter the criteria for rejecting the null hypothesis. A common choice for \(\alpha\) is 0.05, meaning there is a 5% chance we might incorrectly reject a true null hypothesis. In our example, we used \(\alpha = 0.05\) to test the DMV's claim about the pass rate for teens. This choice balances the risk of making a Type I error (false positive) while still allowing for meaningful conclusions.
Choosing the right significance level is vital. Lower levels like 0.01 mean stricter criteria for significance, reducing the risk of Type I errors but possibly increasing Type II errors. Conversely, higher levels, like 0.10, reduce Type II errors but may accept more false positives.
Critical Value
The critical value in hypothesis testing is a key indicator used to determine the threshold at which a test statistic is considered significant. It delineates the boundary beyond which we decide to reject the null hypothesis in favor of the alternative hypothesis.
For a two-tailed test, the critical value is like a gatekeeper on both ends of the distribution. When we choose an \(\alpha = 0.05\), our critical values mark the areas in the tails of the distribution that comprise 5% of the area. These boundaries are often denoted by \(\z^{*}\). In the DMV case, the critical values are approximately ±1.96.
If the calculated test statistic falls beyond these critical values, it suggests the observed effect is statistically significant. However, if the test statistic lies within these bounds, we do not have enough evidence to reject the null hypothesis. This way, the critical value plays an explicit role in deciding whether observed results can reasonably occur if the null hypothesis is true.
Test Statistic
The test statistic is a standardized value used to decide whether to reject the null hypothesis. In our context of proportions, it’s calculated using the formula for the z-score of a sample proportion:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]
Here, \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size. The test statistic tells us how far our sample proportion \(\hat{p}\) deviates from the hypothesized proportion \(p_0\), scaled by the standard error.
In the DMV example, the test statistic was calculated to be 2.575. This tells us our observed sample result (68.8%) is 2.575 standard errors above the hypothesized 60% pass rate. The larger the absolute value of this statistic, the further away our sample finding is from the null hypothesis expectation, which suggests stronger evidence against the null hypothesis.

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Most popular questions from this chapter

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