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Chapter 15: Problem 4

A recent article reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At the .01 significance level, can we conclude that a larger proportion of students at your school have jobs?

Short Answer

Expert verified
No, there is not enough evidence at the 0.01 significance level to conclude that a larger proportion of students at your school have jobs.

Step by step solution

01

State the hypotheses

We start by setting up our null and alternative hypotheses. The null hypothesis (H_0) states that the proportion of students with jobs from your school is equal to the national proportion, which is 1/3 or 0.333. The alternative hypothesis (H_a) states that the proportion of students with jobs from your school is greater than the national proportion.\[ H_0: p = 0.333 \]\[ H_a: p > 0.333 \]
02

Set significance level and find critical value

The significance level is given as 0.01. Since this is a one-tailed test, we find the critical value for Z at 0.01 significance level in the Z-table, which is 2.33. This means that any Z value greater than 2.33 will lead us to reject the null hypothesis.
03

Calculate the test statistic

First, let's calculate the sample proportion \(\hat{p}\) which is 80 out of 200 graduates:\[\hat{p} = \frac{80}{200} = 0.4\]The test statistic is calculated using the formula:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]Where \(\hat{p}\) is the sample proportion, \(p_0 = 0.333\) is the population proportion, and \(n = 200\) is the sample size.
04

Fill in values to compute Z

Substitute the known values into the test statistic formula:\[Z = \frac{0.4 - 0.333}{\sqrt{\frac{0.333(1-0.333)}{200}}}\]This simplifies to:\[Z \approx \frac{0.067}{\sqrt{\frac{0.333 \times 0.667}{200}}}\]Calculate the value inside the square root and the Z value.
05

Comparison and conclusion

Compute the Z value:\[Z \approx \frac{0.067}{0.0324} \approx 2.07\]Since 2.07 is less than the critical value of 2.33, we do not reject the null hypothesis. This means there is not enough evidence at the 0.01 significance level to conclude that a larger proportion of students at your school have jobs.

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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 foundational concept. It represents a statement of no effect or no difference. In simpler terms, it's what we assume to be true until we have evidence to reject it. For instance, when we say "there鈥檚 no difference in employment rates," we are framing our null hypothesis. In our example, the null hypothesis (\(H_0\)) is that the proportion of students with jobs from the school (\(p\)) is equal to the national job proportion, which is 1/3 or 0.333. This traditional approach sets a baseline; anything beyond this, supported by the data, can challenge the status quo.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis suggests that there is a significant effect or a difference. It's all about the idea that something is 'going on'. For our college graduates, the alternative hypothesis (\(H_a\)) proposes that the job proportion for the school is greater than the national average of 0.333. This is where the real investigation begins, as it challenges the initial presumption set by the null. The alternative hypothesis is what researchers are usually trying to prove, creating a dynamic tension between 'what is' and 'what might be'.
Significance Level
Significance level is a critical part of hypothesis testing. It defines the probability of rejecting the null hypothesis when, in fact, it is true. Commonly denoted as \(\alpha\), it is a threshold that determines how strong the evidence must be before we reject the null hypothesis. In this exercise, the significance level is set at 0.01, meaning there's a 1% risk of concluding there's a difference when there isn't. This low threshold suggests a very cautious approach. A lower significance level means the results must be robust to suggest any valid challenge to the null hypothesis.
Z-Test
The Z-test is a statistical test that is used to determine whether there is a significant difference between sample and population means or proportions. When dealing with large samples, the Z-test becomes a powerful tool because of the Central Limit Theorem, which allows for normal distribution approximation. In our example, we utilize the Z-test to assess if the proportion of employed students from the school is significantly greater than the national average. It involves calculating a Z statistic, which is then compared to a critical value derived from the significance level. This determines if the null hypothesis can be rejected.
Proportion Testing
Proportion testing refers to methods used to determine if there is a statistically significant difference between two proportions. This is essential when evaluating categorical data, like "employed" or "unemployed". In the exercise, proportion testing is applied to compare the employment rate of graduates from the school to the national rate. We calculate the sample proportion (\(\hat{p}\)) and then use it to conduct a Z-test. The test tells us if the observed sample result deviates enough from the population proportion to suggest a true difference, beyond random chance. It's especially useful when assessing outcomes as dichotomous variables (such as 'yes' or 'no' scenarios).

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Most popular questions from this chapter

GfK Research North America conducted identical surveys 5 years apart. One question asked of women was "Are most men basically kind, gentle, and thoughtful?" The earlier survey revealed that, of the 3,000 women surveyed, 2,010 said that they were. The later survey revealed 1,530 of the 3,000 women thought that men were kind, gentle, and thoughtful. At the .05 level, can we conclude that women think men are less kind, gentle, and thoughtful in the later survey compared with the earlier one?

The U.S. Department of Transportation estimates that \(10 \%\) of Americans carpool. Does that imply that \(10 \%\) of cars will have two or more occupants? A sample of 300 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 63 had two or more occupants. At the .01 significance level, can we conclude that \(10 \%\) of cars traveling on the New Jersey Turnpike have two or more occupants?

Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 300 days. The sample information is reported below. To interpret, there were 50 days on which no credit applications were received, 77 days on which only one application was received, and so on. Would it be reasonable to conclude that the population distribution is Poisson with a mean of 2.0 ? Use the .05 significance level. (Hint: To find the expected frequencies, use the Poisson distribution with a mean of \(2.0 .\) Find the probability of exactly one success given a Poisson distribution with a mean of 2.0. Multiply this probability by 300 to find the expected frequency for the number of days on which there was exactly one application. Determine the expected frequency for the other days in a similar manner.)

Each month the National Association of Purchasing Managers surveys purchasing managers and publishes the NAPM index. One of the questions asked on the survey is: Do you think the economy is contracting? Last month, of the 300 responding managers, 160 answered yes to the question. This month, 170 of the 290 managers indicated they felt the economy was contracting. At the . 05 significance level, can we conclude that a larger proportion of the purchasing managers believe the economy is contracting this month?

Research in the gaming industry showed that \(10 \%\) of all slot machines in the United States stop working each year. Short's Game Arcade has 60 slot machines and only 3 failed last year. At the .05 significance level, test whether these data contradict the research report. a. Why can you use a z-statistic as the test statistic? b. State the null and alternate hypotheses. c. Evaluate the test statistic and make the decision. d. What is the \(p\) -value and what does that imply?
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