/*! 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 18 Problems 17 and 18 illustrate th... [FREE SOLUTION] | 91影视

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Problems 17 and 18 illustrate the use of the sign test to test hypotheses regarding a population proportion. The only requirement for the sign test is that our sample be obtained randomly. When dealing with nominal data, we can identify a characteristic of interest and then determine whether each individual in the sample possesses this characteristic. Under the null hypothesis in the sign test, we expect that half of the data will result in minus signs and half in plus signs. If we let a plus sign indicate the presence of the characteristic (and a minus sign indicate the absence), we expect half of our sample to possess the characteristic while the other half will not. Letting \(p\) represent the proportion of the population that possesses the characteristic, our null hypothesis will be \(H_{0}: p=0.5 .\) Use the sign test for Problems 17 and 18 , following the sign convention indicated previously. Trusting the Press In a study of 2302 U.S. adults surveyed online by Harris Interactive 1243 respondents indicated that they tend to not trust the press. Using an \(\alpha=0.05\) level of significance, does this indicate that more than half of U.S. adults tend to not trust the press?

Short Answer

Expert verified
Reject H_{0} if X exceeds the critical value.

Step by step solution

01

State the Hypotheses

Under the null hypothesis, we expect half of the data to result in plus signs and half in minus signs. Here, let a plus sign indicate the presence of the characteristic (not trusting the press). Hence, we set up the hypotheses as follows: Null hypothesis (H_{0}): p = 0.5 Alternative hypothesis (H_{a}): p > 0.5.
02

Determine the Sample Statistics

Identify the sample size ( n ) and the number of plus signs. Here, the sample size n = 2302 U.S. adults and the number of plus signs (respondents not trusting the press) = 1243.
03

Calculate the Test Statistic

The test statistic for the sign test is the number of plus signs. In this case, the test statistic X = 1243 .
04

Determine the Critical Value

For a one-tailed test at the 伪 = 0.05 significance level with n = 2302 , use the binomial distribution to find the critical value for X . Lookup or calculate the critical value for a binomial distribution with parameters p = 0.5 and n = 2302.
05

Make the Decision

Compare the test statistic X to the critical value. If X exceeds the critical value, reject the null hypothesis H_{0}. Otherwise, do not reject H_{0}.
06

Conclusion

Based on the result of the comparison between X and the critical value, conclude whether there is significant evidence at the 伪 = 0.05 level to suggest that more than half of U.S. adults tend to not trust the press.

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Key Concepts

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

hypothesis testing
Hypothesis testing is a systematic method used to test assumptions (or hypotheses) about a population parameter. In this exercise, we look at the sign test to determine whether more than half of U.S. adults tend to not trust the press.
The process involves the following steps:
  • **State the Hypotheses**: Set up a null hypothesis (H鈧) and an alternative hypothesis (H鈧) to test your claim. For example, H鈧: p = 0.5 and H鈧: p > 0.5.
  • **Determine Sample Statistics**: Collect data and identify important metrics (such as the number of plus signs).
  • **Calculate Test Statistic**: Use the sign test to compute the number of plus signs, which is our test statistic.
  • **Determine Critical Value**: Refer to statistical tables or use software to find the critical value for your specific test parameters.
  • **Make a Decision**: Compare your test statistic with the critical value to determine whether to reject or fail to reject the null hypothesis.
Based on this exercise, hypotheses about a proportion are tested using the sign test, simplifying the analysis by counting the number of signs (plus or minus) in the data.
population proportion
Population proportion refers to the ratio of individuals in a population who have a certain characteristic. For example, if we want to understand how many U.S. adults do not trust the press, the population proportion (denoted as **p**) would be the ratio of adults who do not trust the press to the total number of surveyed adults.
Here鈥檚 how you can approach understanding population proportion in hypothesis testing:
  • **Characteristic Identification**: Identify the characteristic of interest鈥攍ike 'not trusting the press' in our example.
  • **Null Hypothesis (H鈧)**: Typically, in a sign test context, we assume that 50% (or 0.5) of the population has the characteristic, so H鈧: p = 0.5.
  • **Alternative Hypothesis (H鈧)**: The alternative might propose that more than 50% have the characteristic, as structured H鈧: p > 0.5.
  • **Sample Proportion**: Calculate the sample proportion by dividing the number of people with the characteristic by the total sample size.
In this study, out of 2302 surveyed, 1243 did not trust the press, suggesting a sample proportion. Our task is to identify if this sample proportion provides evidence that more than half of U.S. adults distrust the press.
binomial distribution
The binomial distribution is a probability distribution that summarizes the likelihood of a value taking on one of two independent states across a set number of trials or observations. It鈥檚 very useful in situations like this exercise, where outcomes are binary - either a plus or a minus.
Here's how the binomial distribution connects to the sign test:
  • **Parameters**: We define the binomial distribution with two parameters: **n** (the number of trials or sample size) and **p** (the probability of success on each trial). In our task, n = 2302 and p = 0.5.
  • **Calculating Probabilities**: The binomial distribution helps compute the probability of observing a certain number of successes (plus signs; not trusting the press).
  • **Critical Value**: To use the binomial distribution for hypothesis testing, locate the critical value for your significance level (alpha), often using statistical tables or software.
  • **Decision Making**: Compare the observed test statistic against the critical value derived from the binomial distribution to accept or reject H鈧.
For this exercise, we use the binomial distribution to determine whether the number of respondents not trusting the press (test statistic X = 1243) is significantly different from what would be expected (1151) under H鈧.

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Most popular questions from this chapter

Zillow.com allows people to research real estate listings. One feature is the "Zestimate," which represents the price Zillow believes the property should sell for. If the Zestimate is accurate, the median difference between the actual selling price and the Zestimate should be zero. The following data represent the difference between the actual selling price and Zestimate for a random sample of 10 real estate transactions. Do the data suggest the Zestimate is inaccurate? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{rrrrr} \hline 2500 & -500 & -79,500 & -13,500 & -500 \\ \hline-1000 & 14,500 & 3,500 & -12,500 & -6000 \\ \hline \end{array} $$

Acid Rain In \(2002,\) the median pH level of the rain in Glacier National Park, Montana, was \(5.25 .\) A biologist thinks that the acidity of rain has decreased since then, which would suggest the pH has increased. She obtains a random sample of 15 rain dates in 2015 and obtains the following data: $$ \begin{array}{lllll} \hline 5.31 & 5.19 & 5.55 & 5.38 & 5.37 \\ \hline 5.19 & 5.26 & 5.29 & 5.27 & 5.19 \\ \hline 5.27 & 5.36 & 5.22 & 5.28 & 5.24 \\ \hline \end{array} $$ Test the hypothesis that the median pH level has increased from 5.25 at the \(\alpha=0.05\) level of significance.

Every year Money magazine publishes its list of top places to live. The following data represent a list of top places to live for a recent year, along with the median family income and median commute time. $$ \begin{array}{lcc} \text { City } & \begin{array}{c} \text { Family Income } \\ (\$ \mathbf{1 0 0 0 s}) \end{array} & \begin{array}{c} \text { Commute Time } \\ \text { (minutes) } \end{array} \\ \hline \text { Woodridge, Illinois } & 83 & 27.1 \\ \hline \text { Urbandale, Idaho } & 82 & 17.0 \\ \hline \text { La Palma, California } & 86 & 26.9 \\ \hline \text { Friendswood, Texas } & 90 & 26.0 \\ \hline \text { Suwanee, Georgia } & 101 & 32.1 \\ \hline \text { Somers, Connecticut } & 83 & 22.6 \\ \hline \end{array} $$ (a) Does a positive association exist between income and commute time at the \(\alpha=0.10\) level of significance? (b) Draw a scatter diagram to support your conclusion.

Why is it appropriate to call nonparametric statistical procedures distribution-free procedures?

Reaction-Time Experiment Researchers at the University of Mississippi wanted to learn the reaction times of students to different stimuli. In the following data, the reaction times for subjects were measured after they received a simple stimulus and a go/no-go stimulus. The simple stimulus was an auditory cue, and the time from when the cue was given to when the student reacted was measured. The go/no-go stimulus required the student to respond to a particular stimulus and not respond to other stimuli. Again, the reaction time was measured. The following data were obtained: $$ \begin{array}{ccc|ccc} \begin{array}{l} \text { Subject } \\ \text { Number } \end{array} & \text { Simple } & \text { Go/No-Go } & \begin{array}{l} \text { Subject } \\ \text { Number } \end{array} & \text { Simple } & \text { Go/No-Go } \\ \hline 1 & 0.220 & 0.375 & 16 & 0.498 & 0.565 \\ \hline 2 & 0.430 & 1.759 & 17 & 0.262 & 0.402 \\ \hline 3 & 0.338 & 0.652 & 18 & 0.620 & 0.643 \\ \hline 4 & 0.266 & 0.467 & 19 & 0.300 & 0.351 \\ \hline 5 & 0.381 & 0.651 & 20 & 0.424 & 0.380 \\ \hline 6 & 0.738 & 0.442 & 21 & 0.478 & 0.434 \\ \hline 7 & 0.885 & 1.246 & 22 & 0.305 & 0.452 \\ \hline 8 & 0.683 & 0.224 & 23 & 0.281 & 0.745 \\ \hline 9 & 0.250 & 0.654 & 24 & 0.291 & 0.290 \\ \hline 10 & 0.255 & 0.442 & 25 & 0.453 & 0.790 \\ \hline 11 & 0.198 & 0.347 & 26 & 0.376 & 0.792 \\ \hline 12 & 0.352 & 0.698 & 27 & 0.328 & 0.613 \\ \hline 13 & 0.285 & 0.803 & 28 & 0.952 & 1.179 \\ \hline 14 & 0.259 & 0.488 & 29 & 0.355 & 0.636 \\ \hline 15 & 0.200 & 0.281 & 30 & 0.368 & 0.391 \\ \hline \end{array} $$ The researchers used Minitab to test whether the simple stimulus had a lower reaction time than the go/no-go stimulus. The results of the analysis are as follows: (a) State the null and alternative hypotheses. (b) Is the median reaction time for the go/no-go stimulus higher than the median reaction time for the simple stimulus? Use the \(\alpha=0.05\) level of significance. Why?

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