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Does College Pay Off? An article in Time describing various aspects of American life indicated that higher educational achievement paid off! College grads work 7.4 hours per day, fewer than those with less than a college education. \(^{2}\) Suppose that the average work day for a random sample of \(n=100\) individuals who had less than a four-year college education was calculated to be \(\bar{x}=7.9\) hours with a standard deviation of \(s=1.9\) hours. a. Use the \(p\) -value approach to test the hypothesis that the average number of hours worked by individuals having less than a college degree is greater than individuals having a college degree. At what level can you reject \(H_{0} ?\) b. If you were a college graduate, how would you state your conclusion to put yourself in the best possible light? c. If you were not a college graduate, how might you state your conclusion?

Short Answer

Expert verified
Answer: The data supports the assertion that individuals without a college degree tend to work more hours per day than those with a college degree, as we rejected the null hypothesis at a 0.05 significance level.

Step by step solution

01

State the null and alternative hypotheses

\(H_0: \mu \le 7.4\) -> The average workday hours for individuals without a college degree is less than or equal to those with a degree. \(H_1: \mu > 7.4\) -> The average workday hours for individuals without a college degree is greater than those with a degree.
02

Determine the significance level

Let's use a significance level (\(\alpha\)) of 0.05 as it is commonly used. If the p-value is less than or equal to \(\alpha\), we reject the null hypothesis.
03

Calculate the test statistic

Test statistic for a one-sample t-test: \(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\) Let's plug the values to calculate the test statistic: \(t = \frac{7.9 - 7.4}{\frac{1.9}{\sqrt{100}}} = \frac{0.5}{\frac{1.9}{10}} = \frac{0.5}{0.19} = 2.63\)
04

Calculate the p-value

Using a t-distribution table or calculator, find the p-value corresponding to the given test statistic (\(t=2.63\)) with 99 degrees of freedom (since \(n=100\) individuals). The resulting p-value is 0.0049.
05

Compare p-value to the significance level

We have a p-value of 0.0049, which is less than our chosen significance level \(\alpha = 0.05\). Therefore, we reject the null hypothesis.
06

Conclusion

a. We can reject the null hypothesis at a 0.05 significance level. b. If you were a college graduate, you could state that the data supports the assertion that college graduates work fewer hours per day than those without a degree. This implies that college education pays off, rewarding graduates with a more balanced work-life situation. c. If you were not a college graduate, you could state that the data shows that individuals without a college degree tend to work more hours per day than college graduates. This demonstrates that non-graduates show commitment and hard work in their jobs, which could be seen as a strength in the workforce.

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Key Concepts

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

t-test
A t-test is a statistical test used to compare the means of two groups to understand if they are significantly different from each other. When working with more than one mean, a t-test helps to determine if observed differences are based on random chance or true differences in the groups. There are different types of t-tests depending on the data being analyzed:
  • One-sample t-test: Compares the mean of a single group to a known value (like a national average).
  • Independent two-sample t-test: Compares means between two independent groups (like male and female students).
  • Paiired sample t-test: Compares means from two related groups (like pre-test and post-test results).
In the provided exercise, a one-sample t-test is conducted. This test evaluates whether the average work hours of non-degree holders are significantly different from those with college degrees. By calculating a t-statistic, one determines how far the mean work hours of non-degree holders deviate from the average work hours expected (7.4 hours), considering variances due to sample size and variability. The t-test employed in the exercise aids in logically accepting or rejecting hypotheses about the population based on sample data. Each calculated t-value corresponds to a specific p-value that helps in decision-making about hypotheses.
p-value
The p-value is a critical concept in hypothesis testing that helps us decide whether to reject the null hypothesis. It represents the probability of obtaining results at least as extreme as those observed, under the assumption that the null hypothesis is true. In simple terms, it shows us how likely our sample results align with the scenario proposed under the null hypothesis. Here's what different p-values generally imply:
  • A low p-value (typically ≤ 0.05): Indicates strong evidence against the null hypothesis, leading to its rejection.
  • A high p-value (typically > 0.05): Suggests insufficient evidence to reject the null hypothesis.
  • P-value exactly 0.05: On the borderline, further evidence may be considered.
In the context of our exercise, the calculated p-value was 0.0049, significantly lower than the common significance level of 0.05. This small p-value suggests that there is a very low probability that the observed difference in work hours was due to random chance. Hence, the null hypothesis is rejected, reinforcing the idea that non-degree holders work more hours on average compared to college graduates.
null hypothesis
The null hypothesis is a fundamental part of statistical testing, often denoted as \(H_0\). It provides a statement or claim that there is no effect or no difference, serving as a starting point for testing. It is the hypothesis that researchers typically aim to challenge or disprove.In hypothesis testing, we establish the null hypothesis to facilitate a testable statement. If the evidence in the data significantly contradicts the null hypothesis, it leads to its rejection in favor of the alternative hypothesis.In our exercise, the null hypothesis \(H_0\) was defined such that the average workday hours for individuals without a college degree is less than or equal to those with a college degree. It posits that further educational attainment does not lead to fewer working hours, negating the relationship implied by "Does College Pay Off?" By using statistical tests, we assess whether the data collected provides enough evidence to refute \(H_0\). If enough evidence exists, as it did with our significant p-value of 0.0049, the null hypothesis is rejected, suggesting it is plausible that college graduates work fewer hours than their non-degree counterparts.

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Most popular questions from this chapter

Find the \(p\) -value for the following large-sample \(z\) tests: a. A right-tailed test with observed \(z=1.15\) b. A two-tailed test with observed \(z=-2.78\) c. A left-tailed test with observed \(z=-1.81\)

Actinomycin D A biologist hypothesizes that high concentrations of actinomycin D inhibit RNA synthesis in cells and hence the production of proteins as well. An experiment conducted to test this theory compared the RNA synthesis in cells treated with two concentrations of actinomycin D: .6 and .7 microgram per milliliter. Cells treated with the lower concentration (.6) of actinomycin D showed that 55 out of 70 developed normally, whereas only 23 out of 70 appeared to develop normally for the higher concentration (.7). Do these data provide sufficient evidence to indicate a difference between the rates of normal RNA synthesis for cells exposed to the two different concentrations of actinomycin D? a. Find the \(p\) -value for the test. b. If you plan to conduct your test using \(\alpha=.05\), what will be your test conclusions?

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PCBs Polychlorinated biphenyls (PCBs) have been found to be dangerously high in some game birds found along the marshlands of the southeastern coast of the United States. The Federal Drug Administration (FDA) considers a concentration of PCBs higher than 5 parts per million (ppm) in these game birds to be dangerous for human consumption. A sample of 38 game birds produced an average of 7.2 ppm with a standard deviation of \(6.2 \mathrm{ppm}\). Is there sufficient evidence to indicate that the mean ppm of \(\mathrm{PCBs}\) in the population of game birds exceeds the FDA's recommended limit of 5 ppm? Use \(\alpha=.01\)

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