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A recent survey conducted by Lieberman Research Worldwide and Charles Schwab reported that the "High cost of Living" was the top concern that most surprised young adults as they began life on their own. Twenty-six percent reported "High cost of Living" as their top concern. A disbeliever of this information took his own random sample of 500 young adults starting out on their own in an attempt to show that the true percentage for this top concern is actually higher. a. Find the \(p\) -value if 148 of the young adults surveyed put down "High cost of Living" as their top concern. b. Explain why it is important for the level of significance to be established before the sample results are known.

Short Answer

Expert verified
a) The calculated p-value is about 0.08. b) The level of significance should be established before the sample results are known to avoid bias in data interpretation and to manage the rate of type I error, i.e., rejecting a true null hypothesis.

Step by step solution

01

Define null and alternative hypotheses

In this scenario, the null hypothesis \(H_0\) suggests that the actual population proportion \(p\) is equal to the percentage from the original survey, that is \(p = 0.26 \). The alternative hypothesis \(H_a\) says that the population proportion is greater than the percentage from the original survey, that is \(p > 0.26 \)
02

Calculate test statistic

Next, calculate the test statistic using the formula for a one-sample proportion test: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where \[\hat{p}\] is the sample proportion (148 out of 500, or 0.296), \(p_0\) is the population proportion under the null hypothesis (0.26), and \(n\) is the sample size (500). Plugging these values into the formula gives a Z-value of approximately 1.41.
03

Calculate p-value

The p-value is calculated by examining the probability of achieving the observed z-score of 1.41 or higher, under the assumption of the null hypothesis holding. Using standard normal distribution tables, or software, the p-value is found to be approximately 0.08.
04

Explanation for significance level

The significance level, also referred to as the 'alpha', serves as the threshold that our p-value must reach to reject the null hypothesis. Establishing it before the study or observation avoids bias in interpretation and decision making post hoc. It also helps to manage type I error rate, which is the probability of rejecting a true null hypothesis. If it were not set before, it could be manipulated to favor desired findings.

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Key Concepts

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

Significance Level
When conducting a hypothesis test, the **significance level** is a crucial concept to understand. It is denoted by the Greek letter alpha (\( \alpha \)) and represents the threshold of probability at which we will decide whether or not to reject the null hypothesis. Typically, common values for \( \alpha \) are 0.05, 0.01, or 0.10.

The significance level plays two main roles:
  • It is a cutoff for determining what is considered a statistically significant result.
  • It controls the Type I error rate, which is the chance of falsely rejecting the null hypothesis when it is actually true.
Before analyzing the data, researchers choose a significance level to prevent bias. By setting a threshold before testing, the findings remain objective, minimizing the chance of manipulating results to suit a desired narrative.

In the context of our exercise, setting a significance level is crucial to determine if the sample evidence significantly supports that the true proportion is actually higher than the initially reported 26%.
Population Proportion
The **population proportion** refers to the fraction of individuals within a population that have a particular attribute of interest. In statistical hypotheses concerning proportions, we often denote this parameter as \( p \).

For our exercise, the population proportion in question is the percentage of young adults who consider "High cost of Living" as their top concern. The initial survey stated this figure as 26% or 0.26.

Understanding \( p \) is fundamental in hypothesis testing because it sets a baseline for comparison. In hypothesis testing, we often question whether the sample proportion, which we estimate from collected data, provides enough evidence to suggest that \( p \) truly differs from a known or claimed value.
One-Sample Proportion Test
The **one-sample proportion test** is a statistical method used to determine whether there is enough evidence to conclude that a sample proportion is different from a hypothesized population proportion. This test helps in assessing claims or hypotheses about a population's characteristic based on sample data.

For the exercise at hand, we employ this test to see if the true population proportion of young adults concerned about the high cost of living exceeds 0.26, the proportion from the initial survey.

Here's how the test works:
  • First, calculate the sample proportion \( \hat{p} \). In this exercise, it is 148/500 = 0.296.
  • Compute the z-score using the formula:
    \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
    where \( p_0 \) is the population proportion under the null hypothesis, and \( n \) is the sample size.
  • Determine if the calculated z-score lies in the critical region by comparing it to the chosen significance level, thereby assessing whether to reject the null hypothesis.
This test is a powerful tool for situations like this, where we want to assert conclusions about a particular group's sentiment based on a carefully collected sample.
Null and Alternative Hypotheses
Every hypothesis test involves establishing the **null and alternative hypotheses**. This pair of hypotheses are pivotal in determining the direction and conclusions of the test.

The **null hypothesis** (denoted as \( H_0 \)) represents a statement of no effect or no difference. It often conforms to the assumption that any observed effect or difference is due to random chance. For the exercise with young adults, the null hypothesis asserts the population proportion \( p = 0.26 \), aligning with the original survey's result.

Conversely, the **alternative hypothesis** (denoted as \( H_a \)) suggests what we are testing for or trying to prove. In this setting, it claims the population proportion \( p > 0.26 \), indicating a higher concern about the cost of living among young adults than reported.

Choosing these hypotheses is not arbitrary but rather depends on the context and the specific research question. In hypothesis testing, making a clear distinction between these two leads to objective and focused statistical investigations. The outcomes, either staying with \( H_0 \) or shifting to \( H_a \), rely on data-driven evidence. The clarity and objectivity of these premises ensure a scientifically robust approach to exploring societal trends and behaviors.

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Most popular questions from this chapter

You are testing the hypothesis \(p=0.7\) and have decided to reject this hypothesis if after 15 trials you observe 14 or more successes. a. If the null hypothesis is true and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. b. Find the significance level of your test. c. If the true probability of success is \(1 / 2\) and you observe 13 successes, which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. d. Calculate the \(p\) -value for your hypothesis test after 13 successes are observed.

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. The standard deviation has increased from its previous value of 24. b. The standard deviation is no larger than 0.5 oz. c. The standard deviation is not equal to \(10 .\) d. The variance is no less than \(18 .\) e. The variance is different from the value of \(0.025,\) the value called for in the specs.

An April \(21,2009,\) USA Today article titled "On road, it's do as I say, not as I do" reported that \(58 \%\) of U.S. adults speed up to beat a yellow light. Suppose you conduct a survey in your hometown of 150 randomly selected adults and find that 71 out of the 150 admit to speeding up to beat a yellow light. Does your hometown have a lower rate for speeding up to beat a yellow light than the nation as a whole? Use a 0.05 level of significance.

For a chi-square distribution having 35 degrees of freedom, find the area under the curve between \(\chi^{2}(35,0.96)\) and \(\chi^{2}(35,0.15).\)

It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

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