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Chapter 9: Problem 98

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
Conduct a hypothesis test by first stating the null and alternative hypotheses. After collecting data and calculating sample proportions, compute the test statistic (Z) using the sample proportion, null hypothesis proportion, and sample size. Lastly, choose a significance level and based on the value of the test statistic, decide whether or not to reject the null hypothesis. Depending on this decision, the percentage of students with off campus jobs at your school may or may not be significantly different from \(65\% \).

Step by step solution

01

Formulate the Hypotheses

First, formulate null (H0) and alternative hypothesis (H1). The null hypothesis will assert that the proportion of students working off-campus in your school is \(65\% \), while the alternative hypothesis will assert a different proportion. Mathematically, these can be expressed as follows: H0: p = 0.65 H1: p 鈮� 0.65
02

Collection and Analysis of Data

Collect employment data from 40 students in the school and then calculate the proportion (\( \hat{p} \)) of students who hold off-campus jobs.
03

Compute the Test Statistic

Compute the test statistic using the formula \(Z = \frac{(\hat{p} - p)}{\sqrt{ \frac{p(1-p)}{n} }} \) where:- \( \hat{p} \) is the proportion of students in your sample working off-campus- \( p \) is the conjectured proportion in the null hypothesis (0.65)- \( n \) is the number of students sampled (40)
04

Choose the Significance Level and Conclude the Test

You have to choose a significance level (let's say \( \alpha = 0.05 \)). The decision to reject or fail to reject the null hypothesis will then be based on whether the computed test statistic falls within the acceptance region or the rejection region. You'd need to refer to the standard normal (Z) distribution table for obtaining the critical values. If the calculated Z value lies beyond the critical Z value (in the tails of the distribution), the null hypothesis is rejected, otherwise it's not.

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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 statement that there is no effect or no difference. It is the default assumption that there is nothing unusual happening. In our exercise, the null hypothesis asserts that the proportion of students working off-campus is exactly 65%. This can be mathematically expressed as:
  • Null Hypothesis (\(H_0\)): \(p = 0.65\)
The null hypothesis is what we try to disprove or reject, posing it against the alternative hypothesis. It's like claiming, "Everything is normal!" and seeking data evidence to prove otherwise.
Alternative Hypothesis
The alternative hypothesis is the statement that indicates the presence of an effect or difference. It goes against the null hypothesis, suggesting that something is indeed happening. In our scenario, the alternative hypothesis suggests that the percentage of students holding off-campus jobs is not 65%. We express this mathematically as:
  • Alternative Hypothesis (\(H_1\)): \(p eq 0.65\)
This hypothesis is crucial because it helps determine whether there is enough statistical evidence to reject our initial assumption (the null hypothesis). It opens the door to discovering new findings.
Significance Level
The significance level, often denoted by \(\alpha\), determines how strict our criterion is for rejecting the null hypothesis. Commonly set at 0.05, it represents the probability of rejecting the null hypothesis when it is actually true. This is known as a Type I error:
  • If \(\alpha = 0.05\), we are accepting a 5% chance of incorrectly rejecting the null hypothesis.
Selecting a significance level is crucial, as it impacts how conservative or liberal the test decision is. Once set, we compare the test statistic to appropriate critical values to decide if the null hypothesis can be rejected.
Test Statistic
The test statistic is a standardized value we calculate from sample data during hypothesis testing. It helps determine how far the sample statistic diverges from the null hypothesis. In our exercise, the test statistic is calculated using the following formula:
  • \[ Z = \frac{(\hat{p} - p)}{\sqrt{ \frac{p(1-p)}{n} }} \]
Where:
  • \(\hat{p}\) is the sample proportion (percentage of students with off-campus jobs).
  • \(p\) is the believed proportion according to the null hypothesis.
  • \(n\) is the sample size (in our case, 40 students).
This test statistic is compared against critical values from the standard normal distribution to decide on rejecting or not rejecting the null hypothesis.
Standard Normal Distribution
The standard normal distribution, also known as the Z distribution, is a special normal distribution where the mean is 0 and the standard deviation is 1. It is commonly used in statistics, particularly for hypothesis testing. When computing a test statistic, like a Z-score from our sample, we use the standard normal distribution to find critical values and p-values:
  • Z-scores tell us how many standard deviations our sample mean is away from the population mean.
  • These scores are compared to a Z-table or standard normal distribution table.
  • Each Z-score corresponds to a point on the distribution, helping determine probabilities and decision criteria in hypothesis testing.
Effectively, it acts as a universal ruler to measure how extreme our sample data is relative to the null hypothesis.

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Most popular questions from this chapter

Briefly explain the conditions that must hold true to use the \(t\) distribution to make a test of hypothesis about the population mean.

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed, and they vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.035\) inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, \(\mu=36\) inches, against the alternative hypothesis, \(\mu \neq 36\) inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.015\) inches. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be \(.02 ?\) What if the maximum probability of a Type I error is .10? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02 .\) Does the machine need to be adjusted? What if \(\alpha=.10\) ?

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 the \(2.5 \%\) 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 \(\alpha=.01 ?\) What if \(\alpha=.05\) ?

Consider \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b, Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of 7 . Using \(\alpha=.025\), would you reject the null hypothesis?

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and readjust it if she chooses the maximum probability of a Type I error to be \(.01 ?\) What if the maximum probability of a Type I error is .05? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(\alpha=.05 ?\)
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