/*! 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 49 Teen drivers A state鈥檚 Divisio... [FREE SOLUTION] | 91影视

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Teen drivers A state鈥檚 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 this good evidence that the DMV鈥檚 claim is incorrect? Carry out a test at the A 0.05 significance level to help answer this question.

Short Answer

Expert verified
The evidence suggests the DMV's claim is incorrect at the 0.05 level.

Step by step solution

01

Set up hypotheses

We begin by setting up the null and alternative hypotheses. The null hypothesis (\( H_0 \)) states that the true proportion of teens who pass the test on their first try is equal to the DMV's claim: \( p = 0.60 \). The alternative hypothesis (\( H_a \)) states that the true proportion is different from the claim: \( p eq 0.60 \).
02

Calculate sample proportion

Next, we calculate the sample proportion (\( \hat{p} \)) of teens who passed the test on their first try. \( \hat{p} = \frac{86}{125} = 0.688 \).
03

Check conditions for normal approximation

To perform a significance test using a normal approximation, we check that the sample size is large enough. The conditions are \( n \cdot p_0 \geq 10 \) and \( n \cdot (1 - p_0) \geq 10 \), where \( p_0 = 0.60 \). \( 125 \times 0.60 = 75 \) and \( 125 \times 0.40 = 50 \), both greater than 10.
04

Compute test statistic

Calculate the z-test statistic using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]Substituting our values: \( z = \frac{0.688 - 0.60}{\sqrt{\frac{0.60 \times 0.40}{125}}} \), we find \( z \approx 2.247 \).
05

Determine p-value

Using the standard normal distribution, find the p-value for the calculated z-score. Since this is a two-tailed test, we consider both tails: \( p = 2 \times P(Z > 2.247) \). Using a standard normal table or calculator, \( P(Z > 2.247) \approx 0.0124 \), so \( p \approx 0.0248 \).
06

Conclusion

Compare the p-value to the significance level \( \alpha = 0.05 \). Since \( p \approx 0.0248 < 0.05 \), we reject the null hypothesis at the 0.05 significance level. There is sufficient 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.

Null Hypothesis
The null hypothesis is a fundamental concept in statistics, often denoted as \( H_0 \). It represents a default position that there is no effect or no difference in the context of your study. In our exercise, the null hypothesis is that the true proportion of teens passing their driving test on the first attempt is exactly as the DMV claims, which is 60%. This means mathematically, we write it as \( p = 0.60 \).
This hypothesis serves as a starting point for testing and is presumed true until evidence suggests otherwise. In hypothesis testing, the aim is not to prove the null hypothesis but to determine whether there is enough evidence to reject it. If the evidence is strong enough, as indicated by the calculated p-value compared to a pre-set significance level, we conclude that the null hypothesis might be incorrect.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis and is denoted as \( H_a \). It is what researchers typically aim to support by conducting their test. It represents a new claim or effect that stands in opposition to the null. In our example, the alternative hypothesis suggests that the proportion of teens passing the test is not equal to 60%. Hence, we write it as \( p eq 0.60 \).
The alternative hypothesis captures any deviation from the null hypothesis and can be directional or non-directional. Here, it is non-directional as we are investigating if the proportion differs in any direction, either greater or lesser, from the claimed 60%. When the p-value from our test is smaller than the significance level, we have enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis.
Significance Level
The significance level, often represented by the Greek letter \( \alpha \), is a key player in hypothesis testing. It refers to the risk level you are willing to accept for incorrectly rejecting a true null hypothesis, also known as a Type I error. Commonly used levels of significance are 0.05, 0.01, and 0.10. In this exercise, a significance level of 0.05 was chosen.
The choice of the significance level reflects how stringent you want to be when making your decision. A smaller \( \alpha \) reduces the risk of Type I error but increases the chance of missing a real effect (Type II error). By setting this threshold before conducting the test, you maintain objectivity and not let the results influence the decision of how much evidence is significant enough.
Z-Test
A Z-Test is a statistical test used to determine if there is a significant difference between sample data and a known population parameter. It is particularly useful when the sample size is large enough, usually \( n > 30 \), to apply the normal approximation due to the Central Limit Theorem. In our case, the sample size of 125 teens certainly allows the use of a Z-Test.
During a Z-Test, you calculate the Z-Score, which is a measure of how many standard deviations an element is from the mean. You then compare this Z-Score to a standard normal distribution to find the p-value. Mathematically, the Z-Score is given by \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \), where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion under the null hypothesis, and \( n \) is the sample size.
For our driving test example, the computed Z-Score guides us in determining the test's statistical significance. If the resulting p-value is less than the significance level, we reject the null hypothesis, suggesting the reality might differ from what was claimed by the DMV.

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Most popular questions from this chapter

The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that (a) z can be used only for large samples. (b) z requires that you know the population standard deviation S. (c) z requires you to regard your data as an SRS from the population. (d) z applies only if the population distribution is perfectly Normal. (e) z can be used only for confidence intervals.

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