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91Ó°ÊÓ

According to the 2010 Census, \(16 \%\) of the people in the United States are of Hispanic or Latino origin. One county supervisor believes her county has a different proportion of Hispanic people than the nation as a whole. She looks at their most recent survey data, which was a random sample of 437 county residents, and found that 44 of those surveyed are of Hispanic origin. a. State the hypotheses. b. Name the model and check appropriate conditions for a hypothesis test. c. Draw and label a sketch, and then calculate the test statistic and P-value. d. State your conclusion.

Short Answer

Expert verified
The hypotheses are \(H_0: p = 0.16\) and \(H_A: p \neq 0.16\). A one-sample Z-test for proportions is used to test the hypotheses. After ensuring the model conditions are met, the test statistic and P-value are calculated. The conclusion is drawn based on the comparison of the P-value and the significance level.

Step by step solution

01

Stating the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of Hispanic people in the county is the same as the national proportion, \(p_0 = 0.16\). The alternative hypothesis (\(H_A\)) is that the proportion in the county is not the same as the national proportion, \(p \neq 0.16\).
02

Naming the model & Checking conditions

The model used is a one-sample Z-test for proportions. We check the randomly sampled condition, the large count condition, and the independent condition must also be met. These conditions are: \n1. Randomness: The survey data was randomly sampled. \n2. At least 10 success and 10 failures: This means we need to have at least 10 Hispanic and 10 non-Hispanic people in both our sample and the population. With a population proportion of 0.16 and a sample of size 437, we have 70 successes and 367 failures in the population and 44 successes and 393 failures in the sample, which are more than 10.\n3. Independence: As the sample size (437) is less than 10% of the population size, independence is assumed.
03

Calculating the Test Statistic and P-value

The test statistic, Z, is calculated by subtracting the hypothesized proportion from the sample proportion, and dividing by the standard error of the sample proportion. \nAlso, the P-value associated with this Z-score is found using a Z-table. It gives the probability of getting a result as extreme as, or more extreme than the observed data, under the null hypothesis.
04

Drawing the Conclusion based on P-value

If the P-value is significant (typically, P-value ≤ 0.05), then we reject the null hypothesis in favour of the alternative hypothesis. If the P-value is not significant (P-value > 0.05), we fail to reject the null hypothesis. We compare the calculated P-value with the significance level, and based on this make our conclusion about the hypotheses.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a fundamental concept. It is essentially a default position that indicates no effect or no difference exists. In this exercise, the null hypothesis, noted as \(H_0\), claims that the proportion of Hispanic people in the county is the same as the national level. This translates mathematically to saying \(p_0 = 0.16\), where 0.16 refers to 16% of the population being of Hispanic or Latino origin.
The null hypothesis serves as a starting point for examining evidence against it. By accepting or rejecting it, we can make informed decisions or conclusions based on data at hand. The goal in this context is to use statistical methods to determine whether there is sufficient evidence to refute the null hypothesis.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis offers an assertion that there is an effect or difference present. In this context, it specifies that the proportion of Hispanic people in the county is different from 16%. We represent this as the alternative hypothesis, \(H_A\), mathematically expressed as \(p eq 0.16\). This could indicate either a greater or lesser proportion than the national average.
The alternative hypothesis is critical because it presents the possibility we are seeking to find evidence for. If strong enough evidence is gathered in favor of the alternative hypothesis, it can lead us to reject the null, suggesting a significant difference exists. This forms the basis for making inferential conclusions from sample data.
Z-test for Proportions
The Z-test for proportions is a statistical method used to determine if there is a significant difference between an observed sample proportion and a theoretical population proportion. It's used here to assess whether the county's proportion of Hispanic residents significantly differs from the national percentage.
Before using the Z-test, certain assumptions or conditions must be satisfied:
  • Randomness: The sample must be randomly selected, which improves the reliability of the statistical inference.
  • Large Sample Size: To meet this condition, each group (Hispanic and non-Hispanic) should have at least 10 individuals in the sample.
  • Independence: The sample size should be less than 10% of the total population, ensuring independence between the sampled individuals.
Once these conditions are met, the test statistic, \(Z\), is computed using:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size.
P-value
The P-value is a crucial element in hypothesis testing. It quantifies the evidence against the null hypothesis presented by the observed data. A low P-value suggests that the observed data are unlikely under the null hypothesis, thus providing evidence in support of the alternative hypothesis.
In a Z-test for proportions, the P-value is derived from the test statistic. By looking up the computed Z-value in a Z-table, the P-value indicates the probability of observing a result at least as extreme as the sample's, assuming the null hypothesis is true. Generally, a significance level (often \(\alpha = 0.05\)) sets the threshold for determining the significance of results.
If the computed P-value is less than or equal to this significance level, it provides strong grounds to reject the null hypothesis, pointing towards acceptance of the alternative hypothesis. However, a higher P-value implies insufficient evidence to discard the null hypothesis. This decision-making process is pivotal in assessing whether the differences in observed and expected proportions are statistically significant.

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Most popular questions from this chapter

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.

The nutrition lab in Chapter 17 ?, Exercise 38 ? tested 40 hot dogs to see if their mean sodium content was less than the 325-mg upper limit set by regulations for "reduced sodium" franks. The mean sodium content for the sample was \(322.0 \mathrm{mg}\) with a standard deviation of \(18 \mathrm{mg}\). Assume that the assumptions and conditions for the test are met. a. Test the hypothesis that the mean sodium content meets the regulation. b. Will a larger sample size ensure that the regulations are met?

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.

GRE performance again Instead of advertising the percentage of customers who improve by at least 10 points, a manager suggests testing whether the mean score improves at all. For each customer they record the difference in score before and after taking the course (After \(-\) Before). a. State the null and alternative hypotheses. b. The P-value from the test is 0.65 . Does this provide any evidence that their course works? c. From part \(b,\) what can you tell, if anything, about the mean difference in the sample scores?

More ski wax From Chapter 17 ?, Exercise 58 ?, Bjork Larsen was trying to decide whether to use a new racing wax for cross-country skis. He decided that the wax would be worth the price if he could average less than 55 seconds on a course he knew well, so he planned to study the wax by racing on the course 8 times. His 8 race times were \(56.3,65.9,50.5,52.4,46.5,57.8,52.2,\) and 43.2 seconds. Should he buy the wax? Explain by performing an appropriate hypothesis test.

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