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In March 2001, a Gallup poll asked, "How would you rate the overall quality of the environment in this country today- -as excellent, good, fair or poor?" Of 1060 adults nationwide, 46\% gave a rating of excellent or good. Is this convincing evidence that a majority of the nation's adults think the quality of the environment is fair or poor? Test using \(\alpha=.05 .\)

Short Answer

Expert verified
Yes, the evidence is convincing; a majority rate the environment as fair or poor.

Step by step solution

01

Set Up the Hypotheses

The null hypothesis (\( H_0 \)) is that the proportion of adults rating the environment as fair or poor is \( p = 0.5 \). The alternative hypothesis (\( H_a \)) is that the proportion of adults rating the environment as fair or poor is greater than 0.5. Mathematically, \( H_0: p = 0.5 \) and \( H_a: p > 0.5 \).
02

Calculate the Proportion

According to the poll, 46\% rated the environment as excellent or good. Therefore, the proportion that rated it as fair or poor is \(1 - 0.46 = 0.54\).
03

Determine the Test Statistic

The test statistic is calculated using the formula for the z-test: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] where \( \hat{p} = 0.54 \), \( p_0 = 0.50 \), and \( n = 1060 \).
04

Calculate the Test Statistic Value

Substitute the values into the formula: \[ z = \frac{0.54 - 0.50}{\sqrt{\frac{0.50(1-0.50)}{1060}}} \] This results in \( z \approx 2.05 \).
05

Determine the Critical Value

For a significance level \( \alpha = 0.05 \) for a one-tailed test, the critical value of \( z \) is approximately 1.645.
06

Make the Decision

Since the calculated test statistic \( z \approx 2.05 \) is greater than the critical value of 1.645, we reject the null hypothesis.

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Key Concepts

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

Proportion Test
A proportion test is utilized to decide if a population proportion differs from a specified value. It is a part of hypothesis testing often used when trying to determine if a sample provides enough evidence to conclude that a certain characteristic is true for the entire population. In simpler terms, it helps to check the proportion of a certain trait or opinion in a population, based on a sample.

Looking at our exercise, the poll suggested that 54% of adults rated the environment as fair or poor. This is the proportion of interest. By conducting a proportion test, we aim to decide if this percentage genuinely represents the view of a majority of the population or if it might just be due to random chance.
Null Hypothesis
The null hypothesis is a statement that suggests no effect or no difference in the context of the test. It's the "baseline" assumption that any observed difference is due to random variation.

In this exercise, the null hypothesis ( H_0 ) posits that the proportion of adults who believe the environment is fair or poor is exactly 50%. The reason behind picking this specific value is to serve as a threshold for deciding whether the evidence suggests a shift in opinion. This means if the null hypothesis is accepted, we conclude the proportion of adults holding this view is not a majority, staying at 50%.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis is the statement you aim to support through your data. It counters the null, suggesting there is a real effect or difference. In hypothesis testing, this is what researchers hope to prove.

Here, the alternative hypothesis ( H_a ) declares that the proportion of adults rating the environment as fair or poor is over 50%. Accepting this hypothesis would imply that more than half of the population holds this unfavorable view towards the environment's quality, indicating a significant majority.
Significance Level
The significance level, denoted by \( \alpha \) , indicates the probability of rejecting the null hypothesis when it is actually true. It acts like a threshold, giving a pre-determined risk level for this type of error. It's an essential part of hypothesis testing, guiding decision-making.

For this problem, the significance level is set to 0.05. This standard level means we accept a 5% chance of wrongly concluding that more than half of the population thinks the environment is fair or poor when, in fact, it is not.
  • This needs to be low enough to avoid drawing incorrect conclusions.
  • In many cases, a 5% level is considered a good trade-off between caution and detection.
Z-Test
A Z-test is a type of hypothesis test that determines if there is a significant difference between sample and population means. It is especially useful when the sample is large.

The Z-test utilizes a formula that gives a Z score, showing how many standard deviations our sample proportion is from our hypothesized population proportion. The formula is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] In the context of this problem, we calculated a Z score of approximately 2.05.
  • The Z score is compared to a critical value, which here is 1.645 for a one-tailed test.
  • A Z score higher than this critical value indicates rejecting the null hypothesis.
  • In this example, the Z score suggests sufficient evidence to support the alternative hypothesis.

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Most popular questions from this chapter

Suppose that we are interested in testing the simple null hypothesis \(H_{0}: \theta=\theta_{0}\) versus the simple alternative hypothesis \(H_{a}: \theta=\theta_{a} .\) According to the Neyman-Pearson lemma, the test that maximizes the power at \(\theta_{a}\) has a rejection region determined by $$\frac{L\left(\theta_{0}\right)}{L\left(\theta_{a}\right)}

Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote a random sample from the exponential density with mean \(\theta_{1}\) and let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote an independent random sample from an exponential density with \(\theta_{2}\). a. Find the likelihood ratio criterion for testing \(H_{0}: \theta_{1}=\theta_{2}\) versus \(H_{a}: \theta_{1} \neq \theta_{2}\). b. Show that the test in part (a) is equivalent to an exact \(F\) test [Hint: Transform \(\sum X_{i}\) and \(\sum Y_{j}\) to \(\left.\chi^{2} \text { random variables. }\right]\)

An Article in American Demographics investigated consumer habits at the mall. We tend to spend the most money when shopping on weekends, particularly on Sundays between 4:00 and 6:00 P.M. Wednesday-morning shoppers spend the least. \(^{\star}\) Independent random samples of weekend and weekday shoppers were selected and the amount spent per trip to the mall was recorded as shown in the following table: $$\begin{array}{ll} \hline \text { Weekends } & \text { Weekdays } \\ \hline n_{1}=20 & n_{2}=20 \\ \bar{y}_{1}=\$ 78 & \bar{y}_{2}=\$ 67 \\ s_{1}=\$ 22 & s_{2}=\$ 20 \\ \hline \end{array}$$ a. Is there sufficient evidence to claim that there is a difference in the average amount spent per trip on weekends and weekdays? Use \(\alpha=.05\) b. What is the attained significance level?

In a study to assess various effects of using a female model in automobile advertising, each of 100 male subjects was shown photographs of two automobiles matched for price, color, and size but of different makes. Fifty of the subjects (group A) were shown automobile 1 with a female model and automobile 2 with no model. Both automobiles were shown without the model to the other 50 subjects (group B). In group A, automobile 1 (shown with the model) was judged to be more expensive by 37 subjects. In group \(\mathrm{B}\), automobile 1 was judged to be more expensive by 23 subjects. Do these results indicate that using a female model increases the perceived cost of an automobile? Find the associated \(p\) -value and indicate your conclusion for an \(\alpha=.05\) level test.

The hourly wages in a particular industry are normally distributed with mean \(\$ 13.20\) and standard deviation \(\$ 2.50 .\) A company in this industry employs 40 workers, paying them an average of \(\$ 12.20\) per hour. Can this company be accused of paying substandard wages? Use an \(\alpha=.01\) level test.

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