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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 different 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 sample 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 results of the hypothesis test, particularly the calculated Z value and the associated probability. Based on these, you will either reject or fail to reject the null hypothesis that the proportion of students at your school holding off-campus jobs is 65%.

Step by step solution

01

State the Hypotheses

For a hypothesis test, the first step is to state the null hypothesis (\(H_0\)) and alternative hypothesis (\(H_1\)). In this case, \(H_0\) is that the proportion of students at your school who hold off-campus jobs is 65%, or \(P=0.65\). The alternative hypothesis \(H_1\) is that the proportion is different from 65%, or \(P\neq0.65\).
02

Conduct a Survey

In order to test these hypotheses, data needs to be collected. For this, conduct a survey among 40 students at your school to find out if they hold off-campus jobs.
03

Calculate the Sample Proportion

Once the survey data is collected, calculate the sample proportion. This is done by dividing the number of students who hold off-campus jobs by the total number of students surveyed.
04

Perform a Hypothesis Test

Now, compare the sample proportion with the claimed proportion of 0.65 under the null hypothesis. Since no significance level is given, you are free to choose a standard one, like 0.05. Use a Z-test to determine the test statistic. The Z value will be calculated using the formula: \(Z = (\text{{sample proportion}} - \text{{population proportion}}) / \sqrt{((\text{{population proportion}})(1 - \text{{population proportion}}))/\text{{sample size}}}\)
05

Determine the Decision

Finally, look up the Z value in the Z-table to find the associated probability. If the probability is less than the chosen significance level, reject the null hypothesis. If it is greater, fail to reject the null hypothesis. This will be the conclusion of the hypothesis test.

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Most popular questions from this chapter

A mail-order company claims that at least \(60 \%\) of all orders are mailed within 48 hours. From time to time the quality control department at the company checks if this promise is fulfilled. Recently the quality control department at this company took a sample of 400 orders and found that 208 of them were mailed within 48 hours of the placement of the orders.

Consider the following null and alternative hypotheses: $$ H_{0}: \mu=120 \text { versus } H_{1}: \mu>120 $$ A random sample of 81 observations taken from this population produced a sample mean of \(123.5 .\) The population standard deviation is known to be 15 . a. If this test is made at a \(2.5 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(a=.01 ?\) What if \(\alpha=.05\) ?

For each of the following examples of tests of hypothesis about \(\mu\), show the rejection and nonrejection regions on the \(t\) distribution curve. a. A two-tailed test with \(\alpha=.02\) and \(n=20\) b. A left-tailed test with \(\alpha=.01\) and \(n=16\) c. A right-tailed test with \(\alpha=.05\) and \(n=18\)

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean number of hours spent working per week by college students who hold jobs is different from 20 hours b. To test whether or not a bank's ATM is out of service for an average of more than 10 hours per month c. To test if the mean length of experience of airport security guards is different from 3 years d. To test if the mean credit card debt of college seniors is less than \(\$ 1000\) e. To test if the mean time a customer has to wait on the phone to speak to a representative of a mail-order company about unsatisfactory service is more than 12 minutes

Customers often complain about long waiting times at restaurants before the food is served. A restaurant claims that it serves food to its customers, on average, within 15 minutes after the order is placed. \(A\) local newspaper journalist wanted to check if the restaurant's claim is true. A sample of 36 customers showed that the mean time taken to serve food to them was \(15.75\) minutes with a standard deviation of \(2.4\) minutes. Using the sample mean, the joumalist says that the restaurant's claim is false. Do you think the journalist's conclusion is fair to the restaurant? Use a \(1 \%\) significance level to answer this question.

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