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A study claims that \(65 \%\) of students at all colleges and universities hold off-campus (part-time or full. time) jobs. You want to check if the percentage of students at your school who hold off-campus jobs is dif ferent from \(65 \%\). Briefly explain how you would conduct such a test. Collect data from 40 students at your school on whether or not they hold off-campus jobs. Then, calculate the proportion of students in this sam ple who hold off-campus jobs. Using this information, test the hypothesis. Select your own significance level

Short Answer

Expert verified
The answer depends on the data collected from the 40 students. After collecting the data, calculate the sample proportion and the test statistic (Z-score). If the Z-score lies in the critical region appropriate to the chosen significance level, the null hypothesis is rejected, implying that the proportion of students with off-campus jobs at your school is significantly different from 0.65. If the Z-score doesn't lie in the critical region, it is failed to be rejected, suggesting there's no significant difference.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of students with off-campus jobs at your school is the same as the population proportion, which is 0.65. The alternative hypothesis (\(H_a\)) is that the proportion of students with off-campus jobs at your school is different from 0.65.
02

Collect Data

Randomly select and collect data from 40 students at your school on whether they hold off-campus jobs. Count the number of students who hold off-campus jobs, represented by \(x\), and calculate the sample proportion (\(p\)), which equals \(x/40\).
03

Confirm Conditions

To apply a Z-test, we should fulfill two conditions: the sample size is large enough (i.e., \(n*p >= 10\) and \(n*(1-p) >= 10\)) and the sample is randomly selected, which we assume.
04

Conduct the Hypothesis Test

Calculate the standard deviation of the sample distribution as \(\sqrt{(.65*.35)/40}\). Then, calculate the z-score using the formula \(Z = (p - 0.65)/standard deviation\). Compare the z-score to the critical z-scores for the selected significance level (typically 0.05) to make a decision about the null hypothesis. If the calculated z-score is more than the critical value or less than the negative of the critical value, reject \(H_0\). Otherwise, fail to reject \(H_0\).

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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 fundamental concept. It reflects a statement of "no effect" or "no difference." It is the hypothesis that we assume to be true at the beginning of our test. For the exercise at hand, the null hypothesis (\(H_0\)) states that the proportion of students at your school who hold off-campus jobs is equal to the general population proportion, which is 0.65. By stating this, we're saying that there's no difference in job-holding behavior between students at your school and students nationwide. When conducting experiments, it's important to start by assuming the null hypothesis is true until there is enough evidence against it. This approach helps keep our testing unbiased.
Alternative Hypothesis
The alternative hypothesis is what you are trying to prove in your study. It suggests that there is a difference or change from the condition set by the null hypothesis. In this case, your alternative hypothesis (\(H_a\)) posits that the proportion of students at your school with off-campus jobs deviates from the nationwide rate of 0.65.This hypothesis supports the idea that there's something unique about your school's student population that doesn't match with the general trend. Alternative hypotheses can be one-sided or two-sided. In the exercise, the alternative hypothesis is two-sided, meaning any significant deviation from 0.65—either upward or downward—will provide grounds to reject the null hypothesis.
Z-test
The Z-test is a statistical method used to determine whether there is a significant difference between sample and population means or proportions. It serves as a useful tool in hypothesis testing when the standard deviation is known and the sample size is large enough. For our exercise, we'll use the Z-test to compare the sample proportion of students holding off-campus jobs to the population proportion of 0.65. Before applying the Z-test, ensure two conditions:
  • The sample size must be sufficiently large (both \(n*p\) and \(n*(1-p)\) should be at least 10).
  • The sample should be randomly selected to prevent bias.
Once conditions are met, the z-score is calculated using \[Z = \frac{(p - 0.65)}{\text{standard deviation}}\]where \(p\) is the sample proportion. This score helps in determining whether to retain or reject the null hypothesis.
Sample Proportion
The concept of sample proportion is crucial in statistical hypothesis testing. It is basically the fraction or percentage of a sample that exhibits a particular characteristic. In the exercise, we compute the sample proportion by taking the number of students with off-campus jobs, represented by \(x\), divided by the total sample size, which is 40.Thus, \[p = \frac{x}{40}\]This proportion gives us an estimate of the true population proportion and serves as a basis for further analysis in hypothesis testing.Sample proportion is a straightforward yet critical measure since it provides direct insight into the sample's behavior and how it compares with the expected population proportion. It's one of the stepping stones to perform a Z-test and ultimately draw a conclusion about our hypotheses.

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Most popular questions from this chapter

A past study claimed that adults in America spent an average of 18 hours a week on leisure activi ties. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows \(\begin{array}{lllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9\end{array}\) \(\begin{array}{llllllll}37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.2\end{array}\) Assume that the times spent on leisure activities by all American adults are normally distributed. Us ng a \(10 \%\) significance level, can you conclude that the average amount of time spent by American adults on leisure activities has changed? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections 3.1.1 and 3.2.2 of Chapter 3. Then make the test of hypothesis about \(\mu .\)

A random sample of 500 observations produced a sample proportion equal to \(.38 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.05\). a. \(H_{0}: p=.30\) versus \(H_{1}: p>.30\) b. \(H_{0}: p=.30\) versus \(H_{1}: p \neq .30\)

Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=46, \quad H_{1}: \mu \neq 46, \quad n=40, \quad \bar{x}=49.60, \quad \sigma=9.7\) b. \(H_{0}: \mu=26, \quad H_{1}: \mu<26, \quad n=33, \quad \bar{x}=24.30, \quad \sigma=4.3\) c. \(H_{0}: \mu=18, \quad H_{1}: \mu>18, \quad n=55, \quad \bar{x}=20.50, \quad \sigma=7.8\)

A journalist claims that all adults in her city spend an average of 30 hours or more per month on: general reading, such as newspapers, magazines, novels, and so forth. A recent sample of 25 adults from this city showed that they spend an average of 27 hours per month on general reading. The population of such times is \(\mathrm{kn}\) be normally distributed with the population standard deviation a. Using a \(2.5 \%\) significance level, would you conclude that the mean time spent per month on such reading by all adults in this city is less than 30 hours? Use both procedures - the \(p\) -value approach and the critical value approach. b. Make the test of part a using a \(1 \%\) significance level. Is your decision different from that of part a? Comment on the results of parts a and b.

According to the U.S. Bureau of Labor Statistics, all workers in America who had a bachelor's degree and were employed earned an average of \(\$ 1038\) a week in \(2010 .\) A recent sample of 400 American workers who have a bachelor's degree showed that they earn an average of \(\$ 1060\) per week. Suppose that the population standard deviation of such earnings is \(\$ 160 .\) a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean weekly earning of American workers who have a bachelor's degree is higher than \(\$ 1038 .\) Will you reject the null hypothesis at \(\alpha=.025 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

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