/*! 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 Please provide the following inf... [FREE SOLUTION] | 91Ó°ÊÓ

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Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Generation Gap: Education Education influences attitude and lifestyle. Differences in education are a big factor in the "generation gap." Is the younger generation really better educated? Large surveys of people age 65 and older were taken in \(n_{1}=32\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{1}=15.2 \%\) of the older adults had attended college. Large surveys of young adults (ages \(25-34\) ) were taken in \(n_{2}=35\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{2}=19.7 \%\) of the young adults had attended college. From previous studies, it is known that \(\sigma_{1}=7.2 \%\) and \(\sigma_{2}=5.2 \%\) (Reference: American Generations by S. Mitchell). Does this information indicate that the population mean percentage of young adults who attended college is higher? Use \(\alpha=0.05\)

Short Answer

Expert verified
Reject the null hypothesis; young adults have a significantly higher college attendance rate than older adults at the 0.05 significance level.

Step by step solution

01

Define the Level of Significance and Hypotheses

The level of significance given in the problem is \( \alpha = 0.05 \). We need to set up our null and alternative hypotheses. - Null Hypothesis \( H_0 \): \( \mu_1 \geq \mu_2 \), meaning the percentage of young adults who attended college is not higher than that of older adults.- Alternative Hypothesis \( H_a \): \( \mu_1 < \mu_2 \), indicating the percentage of young adults who attended college is higher.
02

Choose the Sampling Distribution

Here, both population standard deviations (\( \sigma_1 = 7.2 \% \) and \( \sigma_2 = 5.2 \% \)) are known. Thus, we will use the normal distribution (or \(z\)-distribution).Assumptions made: Samples are independent, random, and both sample sizes are fairly large (\(n_1 = 32\) and \(n_2 = 35\)).
03

Compute the Test Statistic

The test statistic is calculated using the formula for the difference between two means:\[z = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Plugging in the values:\[z = \frac{19.7\% - 15.2\%}{\sqrt{\frac{(7.2\%)^2}{32} + \frac{(5.2\%)^2}{35}}} = \frac{4.5\%}{1.634\%} \approx 2.755\]So, the \( z \) value is approximately \( 2.755 \).
04

Calculate the P-value

With a \( z \) value of approximately \( 2.755 \), we check standard normal distribution tables or use software to find the corresponding \( P \)-value. For a \( z \) of 2.755, the \( P \)-value is about 0.003. A sketch of the normal distribution would show a small tail beyond \( z = 2.755 \) representing this \( P \)-value.
05

Decision on Null Hypothesis

Since \( P = 0.003 \) is less than \( \alpha = 0.05 \), we reject the null hypothesis. This indicates that at the 0.05 level of significance, there is enough evidence to support the claim that the proportion of young adults who attended college is higher than that of older adults.
06

Interpret the Results

In the context of the given problem, rejecting the null hypothesis suggests that statistically, there is a significant increase in the percentage of young adults (ages 25-34) who attended college compared to older adults (age 65 and older) across these surveyed U.S. cities. The generational education gap is evident in this data.

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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 starting point. It is the statement you are trying to verify or disprove. It usually represents the "no effect" or "no difference" scenario. For example, if you are testing whether a new teaching method is effective, the null hypothesis might state that this method has no effect on student performance.
In our exercise, the null hypothesis (H_0) is that the percentage of young adults who have attended college is not higher than that of older adults. Symbolically, this is represented as \(\mu_1 \geq \mu_2\). Here, \(\mu_1\) is the mean percentage of older adults who attended college, and \(\mu_2\) is that of young adults.
Understanding the null hypothesis is crucial because it is the foundation against which you compare your alternative hypothesis. If your data lead you to "reject the null hypothesis," it implies there is enough statistical evidence for an alternative explanation or effect.
Alternative Hypothesis
The alternative hypothesis is what you suspect might be true. It is the opposing view to the null hypothesis and is what scientists hope to prove. If a new drug is believed to be more effective than the current standard, the alternative hypothesis would state that the new drug has a greater efficacy.
For the given problem, the alternative hypothesis (H_a) posits that the percentage of young adults who attended college is higher than that of older adults. This can be symbolized as \(\mu_1 < \mu_2\), indicating a generational gap in higher education.
In hypothesis testing, if your statistical analysis suggests you "reject the null hypothesis," it is an indication that your alternative hypothesis might indeed be true. This doesn't absolutely prove the alternative, but it makes a compelling case for considering it as a stronger possibility.
Statistical Significance
Statistical significance helps us decide whether an observed effect or relationship in a study is likely to be genuine or occurred by chance. It's determined by the p-value, which gives us a probability measure. For a result to be statistically significant, its p-value must be less than the chosen level of significance (commonly 0.05). This means there's less than a 5% likelihood that the same result could occur if the null hypothesis were true.
In the exercise, the calculated p-value is 0.003, which is less than the significance level of \(\alpha = 0.05\). This indicates the observed difference, where the percentage of young adults attending college exceeds that of older adults, is statistically significant. There is a strong likelihood that this difference is not due to random variance.
  • Significance Level (\(\alpha\)): Often 0.05 or 0.01 in tests.
  • P-Value: The probability of observing a test statistic at least as extreme as the one observed, assuming the null hypothesis is true.
When a result is statistically significant, it adds weight to the evidence for the alternative hypothesis, guiding scientists and researchers toward further investigations or applications.

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Most popular questions from this chapter

If the \(P\) -value in a statistical test is greater than the level of significance for the test, do we reject or fail to reject \(H_{0} ?\)

Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce \(r_{1}\) successes out of \(n_{1}\), trials for the first population and \(r_{2}\) successes out of \(n_{2}\) trials for the second population. (a) What does the null hypothesis claim about the relationship between the proportions of successes in the two populations? (b) What is the formula for the \(z\) value of the sample test statistic?

Basic Computation: Testing \(p\) A random sample of 30 binomials trials resulted in 12 successes. Test the claim that the population proportion of successes does not equal \(0.50 .\) Use a level of significance of 0.05 (a) Check Requirements Can a normal distribution be used for the \(\hat{p}\) distribution? Explain. (b) State the hypotheses. (c) Compute \(\hat{p}\) and the corresponding standardized sample test statistic. (d) Find the \(P\) -value of the test statistic. (e) Do you reject or fail to reject \(H_{0}\) ? Explain. (f) Interpretation What do the results tell you?

What terminology do we use for the probability of rejecting the null hypothesis when it is true? What symbol do we use for this probability? Is this the probability of a type I or a type II error?

Consider a test for \(\mu\). If the \(P\) -value is such that you can reject \(H_{0}\) at the \(5 \%\) level of significance, can you always reject \(H_{0}\) at the \(1 \%\) level of significance? Explain.

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