/*! 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 103 Use the following information to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 103

Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person鈥檚 nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test. 鈥� 鈥�2011鈥� subscript: 2011 group. 鈥� 鈥�2010鈥� subscript: 2010 group This is: a. a test of two proportions b. a test of two independent means c. a test of a single mean d. a test of matched pairs.

Short Answer

Expert verified
a. a test of two proportions

Step by step solution

01

Identify the Problem Type

Based on the information given, we need to compare the proportion of neuroinvasive West Nile virus cases between two groups (the year 2010 and 2011). This setup fits the criteria for a test of two proportions, as we are comparing two different proportions from two different populations.
02

Choose Hypotheses

Define the null hypothesis as there being no difference in the proportions of neuroinvasive cases between 2010 and 2011. This means \( H_0: p_{2011} = p_{2010} \). The alternative hypothesis is that the 2011 proportion is greater, \( H_a: p_{2011} > p_{2010} \).
03

Calculate Sample Proportions

Compute the proportion of neuroinvasive cases in each year. For 2010, \( \hat{p}_{2010} = \frac{629}{1021} \approx 0.616 \). For 2011, \( \hat{p}_{2011} = \frac{486}{712} \approx 0.683 \).
04

Set Up the Test Statistics

The formula for the test statistic for two proportions is \( Z = \frac{(\hat{p}_{2011} - \hat{p}_{2010})}{\sqrt{ \hat{p}(1 - \hat{p})(\frac{1}{n_{2011}}+\frac{1}{n_{2010}})}} \), where \( \hat{p} \) is the pooled sample proportion. Calculate \( \hat{p} = \frac{629 + 486}{1021 + 712} = \frac{1115}{1733} \approx 0.643 \).
05

Calculate the Test Statistic

Calculate \( Z = \frac{0.683 - 0.616}{\sqrt{0.643 \times (1 - 0.643) \times \left( \frac{1}{712} + \frac{1}{1021} \right)}} \). This simplifies to \( Z \approx 2.52 \).
06

Compare to Critical Value

Using a 1% level of significance, we compare the calculated Z-value to the critical value from a Z-table. At 0.01 significance level, the critical value for a one-tailed test is approximately 2.33. Since 2.52 > 2.33, we reject the null hypothesis.
07

Conclusion

With the test statistic being higher than the critical value, we conclude that the proportion of neuroinvasive West Nile virus cases in 2011 is statistically significantly greater than in 2010.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Test of Two Proportions
When you're faced with the challenge of comparing two sets of data to see if one proportion is greater than the other, you are conducting a test of two proportions. This type of hypothesis test is very useful when dealing with categorical data. Instead of comparing raw numbers, you compare the proportion of occurrences in two different groups.
For instance, in the context of the neuroinvasive West Nile virus, we look at the proportion of cases relative to the total reported in two different years, 2010 and 2011. Here, each year represents a different population, with 2010 being one and 2011 being another. By calculating and comparing these proportions, you can determine if there is a significant difference between the two. This approach is essential for making meaningful inferences based on percentage changes rather than absolute numbers.
Significance Level
In hypothesis testing, the significance level, often denoted as \( \alpha \), is a critical concept. It is the probability of rejecting the null hypothesis when it is actually true. Essentially, it's your threshold for error.
  • A common significance level is 0.05, but in more stringent testing, such as this one for the West Nile virus study, a 1% level of significance is used.
At a 1% level, the researcher is saying they're willing to risk a 1% chance of incorrectly suggesting that the numbers in 2011 are significantly different from those in 2010.
Using a lower significance level, like 1%, implies more confidence in the test's result because it reduces the likelihood of a Type I error (false positive). When conducting such tests, it's essential to select an appropriate significance level as it directly impacts the robustness of your conclusions.
Null and Alternative Hypothesis
The hypothesis formulation is the first and most crucial step in hypothesis testing. The null hypothesis \( (H_0) \) typically states that there is no effect or difference, as a baseline for comparison. In our case, it claims that the proportions of neuroinvasive West Nile virus cases in 2010 and 2011 are equal, \( H_0: p_{2011} = p_{2010} \).
On the other hand, the alternative hypothesis \( (H_a) \) posits that there is a difference. For the West Nile virus scenario, it suggests that the 2011 proportion of cases is greater than that of 2010, \( H_a: p_{2011} > p_{2010} \).
  • The alternative hypothesis guides the direction of the test.
It helps in determining whether to perform a one-tailed or two-tailed test. Since the hypothesis suggests that one specific proportion is greater than the other, we conduct a one-tailed test.
Critical Value
The critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. It's like a dividing line that determines what is statistically significant.
  • In our West Nile virus example, at a 1% significance level, the critical value from the standard normal distribution (Z-table) for a one-tailed test is about 2.33.
The test statistic we calculated was approximately 2.52. Since 2.52 is greater than 2.33, it falls into the rejection region. This tells us that the difference in proportions is statistically significant.
Understanding the critical value helps you set a benchmark for your statistical test, providing a clear point at which a hypothesis is considered refuted by the data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Which distribution (normal or Student's t) would you use for this hypothesis test?

Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. $$\begin{array}{|l|l|l|}\hline \text { Plant Group } & {\text { Sample Mean Height of Plants (inches) }} & {\text { Population Standard Deviation }} \\\ \hline \text { Food } & {16} & {2.5} \\ \hline \text { No food } & {14} & {1.5} \\ \hline\end{array}$$ Draw the graph of the p-value.

Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was \(679. For 23 teenage girls, it was \)559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls.

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.

Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for a. independent group means, population standard deviations, and/or variances known b. independent group means, population standard deviations, and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.