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Pollution A company with a fleet of 150 cars found that the emissions systems of 7 out of the 22 they tested failed to meet pollution control guidelines. Is this strong evidence that more than \(20 \%\) of the fleet might be out of compliance? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Short Answer

Expert verified
Whether the fleet is out of compliance largely or not depends on the p-value. If the p-value is less than 0.05, there is strong evidence that more than 20% of the fleet might be out of compliance. Otherwise, there isn't statistically significant evidence to suggest that.

Step by step solution

01

Hypotheses formulation

The null hypothesis \(H_0\) is that \(20\%\) or fewer of the fleet might be out of compliance i.e., \(p=0.20\). The alternative hypothesis \(H_A\) is that more than \(20\%\) of the fleet might be out of compliance i.e., \(p > 0.20\).
02

Checking Conditions

Before proceeding, it's necessary to check if the conditions for carrying out the hypothesis test are met i.e., Random, Normal, and Independent.\n\na) Random: We’re assuming these cars were chosen in a way that every car had an equal chance to be selected.\nb) Normal: For this condition, we need \(\hat{p}n \geq 10\rightarrow0.20*22\geq10\) and \((1-\hat{p})n\geq 10\rightarrow0.80*22\geq10\).\nThis condition is satisfied.\nc) Independent: We are sampling without replacement, but we can assume independence since 22 cars are less than 10% of the total population of interests i.e., 150 cars. Thus, the condition is satisfied. In conclusion, we're okay to proceed with the Normal model.
03

Calculate Z-score

The next step is to calculate the Z-score which measures how many standard deviations the sample proportion is away from the null hypothesis proportion. The formula for the Z-score is \( Z = \frac{ \hat{p} - p} { \sqrt{\frac{p(1 - p)}{n}} }\). Here,\( \hat{p} = \frac{7}{22} \), \( p = 0.20 \), and \( n = 22 \). Plug these into the formula and calculate the value of Z.
04

Determine the p-value

The p-value is the probability of getting a z-score as extreme as the observed one, or more, assuming \(H_0\) is true. Use a Z-table or statistical software to find the p-value corresponding to the calculated Z-score. If the p-value is less than 0.05, we reject the null hypothesis.
05

Interpretation

If the p-value is less than 0.05, we reject the null hypothesis i.e. there is strong evidence that more than \(20\%\) of the fleet might be out of compliance with pollution control guidelines. If not, we fail to reject the null hypothesis.

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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 the starting point for any hypothesis test. It represents a statement of no effect or no difference, that you want to challenge. In this exercise, the null hypothesis, denoted as \(H_0\), is that 20% or fewer of the cars in the fleet fail to meet pollution control guidelines. This translates to the mathematical expression \(p = 0.20\). The null hypothesis is essentially a benchmark or baseline from which we measure deviation. It remains our assumption until evidence suggests otherwise. We usually retain the null hypothesis unless statistical analysis suggests significant evidence to the contrary.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis proposes a different outcome from the status quo. Here, our alternative hypothesis, represented as \(H_A\), is the claim that more than 20% of the cars might be non-compliant with pollution control requirements. Mathematically, it is expressed as \(p > 0.20\). The alternative hypothesis is what you are trying to provide evidence for through testing. When performing hypothesis testing, the goal is often to find evidence that supports this alternative hypothesis by rejecting the null hypothesis.
Z-score
The Z-score is a statistical measure that shows how far away our sample proportion is from the null hypothesis proportion, expressed in standard deviation units. To compute the Z-score, we use the formula:
\[ Z = \frac{ \hat{p} - p }{ \sqrt{\frac{p(1 - p)}{n}} } \]
Where:
  • \(\hat{p}\) is the sample proportion (i.e., the observed proportion of cars failing, which is \(\frac{7}{22}\)).
  • \(p\) is the null hypothesis proportion (0.20 in this case).
  • \(n\) is the sample size (22).
The computed Z-score helps us understand how extreme our sample result is. The larger the Z-score, the farther our observation is from the null hypothesis, indicating more substantial evidence to reject it.
P-value
The p-value is a crucial measure in hypothesis testing. It quantifies the probability of obtaining a test statistic as extreme as the one observed, given that the null hypothesis is true. In our scenario, a high p-value means that the observed sample result (or one more extreme) might occur under the null hypothesis. A low p-value, typically less than 0.05, suggests that the observed result is quite unlikely under the null hypothesis.
To find the p-value, one usually uses statistical tables or software to see where the observed Z-score lies on the standard normal distribution. A p-value below the chosen significance level (0.05 commonly) leads to rejecting the null hypothesis, indicating stronger evidence for the alternative hypothesis that more than 20% of the fleet fails to meet the guidelines.

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Most popular questions from this chapter

WebZine A magazine is considering the launch of an online edition. The magazine plans to go ahead only if it's convinced that more than \(25 \%\) of current readers would subscribe. The magazine contacted a Simple Random Sample of 500 current subscribers, and 137 of those surveyed expressed interest. What should the company do? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Dice The seller of a loaded die claims that it will favor the outcome \(6 .\) We don't believe that claim, and roll the die 200 times to test an appropriate hypothesis. Our P-value turns out to be 0.03. Which conclusion is appropriate? Explain. a. There's a \(3 \%\) chance that the die is fair. \(\mathrm{b}\). There's a \(97 \%\) chance that the die is fair. c. There's a \(3 \%\) chance that a loaded die could randomly produce the results we observed, so it's reasonable to conclude that the die is fair. d. There's a \(3 \%\) chance that a fair die could randomly produce the results we observed, so it's reasonable to conclude that the die is loaded.

Hypotheses and parameters As in Exercise 3 ?, for each of the following situations, define the parameter and write the null and alternative hypotheses in terms of parameter values. a. Seat-belt compliance in Massachusetts was \(65 \%\) in 2008. The state wants to know if it has changed. b. Last year, a survey found that \(45 \%\) of the employees were willing to pay for on-site day care. The company wants to know if that has changed. c. Regular card customers have a default rate of \(6.7 \% .\) A credit card bank wants to know if that rate is different for their Gold card customers. d. Regular card customers have been with the company for an average of 17.3 months. The credit card bank wants to know if their Gold card customers have been with the company on average the same amount of time.

Obesity 2016 In \(2016,\) the Centers for Disease Control and Prevention reported that \(36.5 \%\) of adults in the United States are obese. A county health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these responses provide strong evidence that the \(36.5 \%\) figure is not accurate for this region? Correct the mistakes you find in a student's attempt to test an appropriate hypothesis. $$ \begin{array}{l} \mathrm{H}_{0}: \hat{p}=0.365 \\ \mathrm{H}_{\mathrm{A}}: \hat{p}<0.365 \\ \mathrm{SRS}, 750 \geq 10 \\ \frac{228}{750}=0.304 ; S D(\hat{p})=\sqrt{\frac{(0.304)(0.696)}{750}}=0.017 \end{array} $$ \(z=\frac{0.304-0.365}{0.017}=-3.588\) $$ \mathrm{P}=P(z>-3.588)=0.9998 $$ There is more than a \(99.98 \%\) chance that the stated percentage is correct for this region.

Write the null and alternative hypotheses you would use to test each of the following situations: a. A governor is concerned about his "negatives"- -the percentage of state residents who express disapproval of his job performance. His political committee pays for a series of TV ads, hoping that they can keep the negatives below \(30 \%\). They will use follow-up polling to assess the ads' effectiveness. b. Is a coin fair? c. Only about \(20 \%\) of people who try to quit smoking succeed. Sellers of a motivational tape claim that listening to the recorded messages can help people quit.

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