/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Perform the following tests of h... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Perform the following tests of hypothesis. a. \(H_{0}: \mu=285, \quad H_{1}: \mu<285, \quad n=55, \bar{x}=267.80, \quad s=42.90, \quad \alpha=.05\) b. \(H_{0}: \mu=10.70, \quad H_{1}: \mu \neq 10.70, \quad n=47, \bar{x}=12.025, \quad s=4.90, \quad \alpha=.01\) c. \(H_{0}: \mu=147,500, \quad H_{1}: \mu<147,500, n=41, \bar{x}=141,812, s=22,972, \alpha=.10\)

Short Answer

Expert verified
a. Reject \(H_{0}\), evidence supports \(H_{1}\). b. Fail to reject \(H_{0}\), insufficient evidence to support \(H_{1}\). c. Reject \(H_{0}\), evidence supports \(H_{1}\). (The actual decision will depend on the calculated test statistics and the specific critical values from the t-distribution table.)

Step by step solution

01

Set up the hypotheses

The null hypothesis (\(H_{0}\)) is stated in each part a, b, and c. The alternative hypothesis (\(H_{1}\)) is stated in each part a, b, and c. The value under the null hypothesis is the assumed population mean, which will be compared to the sample mean.
02

Calculate test statistic

The test statistic is calculated using the formula \(t=\frac{\( \bar{x} - \mu \) }{ \frac{s}{\sqrt{n} }}\). Here, \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, s is the standard deviation of the sample, and n is the sample size. Calculate this for each part.
03

Find critical value(s) and make a decision

Look up the critical value(s) in the t-distribution table using the sample size (n) and the significance level (\( \alpha\)). For a one-tailed test (part a and c), there is one critical value. If the test statistic is less than the critical value, reject the null hypothesis. For a two-tailed test (part b), there are two critical values. If the test statistic is less than the lower critical value or more than the upper critical value, reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

The Null Hypothesis
In hypothesis testing, the null hypothesis (\( H_0 \)) is a statement we assume to be true until evidence suggests otherwise. It typically reflects the idea of "no effect" or "no difference." For example, in a quality control test, one might start with a null hypothesis asserting that the mean of a production process is equal to a target value. This sets a benchmark to test against.

Given the examples in your exercise:
  • In part (a), \( H_0: \mu = 285 \) means you are assuming the population mean is 285.
  • In part (b), \( H_0: \mu = 10.70 \) assumes the population mean is exactly 10.70.
  • In part (c), \( H_0: \mu = 147,500 \) implies that the mean income level is hypothesized to be 147,500.
In each case, the goal is to test whether the sample data presents enough evidence to reject this initial assumption.
Test Statistic
The test statistic is a crucial part of hypothesis testing. It helps to determine how far the sample statistic (like the sample mean) is from the parameter specified in the null hypothesis. The test statistic for a mean in a t-test is calculated by:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]Here's what each symbol represents:
  • \( \bar{x} \) is the sample mean.
  • \( \mu \) is the population mean under the null hypothesis.
  • \( s \) is the sample's standard deviation.
  • \( n \) is the sample size.
This formula essentially measures the difference between the sample mean and the population mean, standardized by the standard error. By plugging in numbers from actual data, you can calculate the exact test statistic value to further evaluate the hypothesis.
T-Distribution
The t-distribution is a probability distribution that is used in hypothesis testing, particularly when dealing with small sample sizes or when the population standard deviation is unknown. Unlike the normal distribution, the t-distribution is broader and has heavier tails. As the sample size increases, it becomes similar to a normal distribution.

In hypothesis testing, if the test statistic falls into the critical region of the t-distribution (determined by the chosen significance level \( \alpha \)), we then have enough evidence to reject the null hypothesis. The critical values from the t-distribution depend on:
  • The degrees of freedom, which is often the sample size minus one (\( n - 1 \)).
  • The significance level (\( \alpha \)), which dictates how much risk of error you're willing to accept.
For example, in part (b) with a two-tailed test of \( \alpha = 0.01 \), you'll find two critical values corresponding to both tails of the distribution. The t-distribution is key because it adjusts for the sample's size and variability when no exact population parameters are known.
Sample Mean
The sample mean (\( \bar{x} \)) is a fundamental concept in statistics as it provides an estimate of the population mean. It is calculated as the average of all observations in a sample. In hypothesis testing, the sample mean helps us understand how close or far the sample observations are on average from the hypothesized population mean.

For example, in part (a), your sample mean is 267.80, a value used to test against the null hypothesis of \( \mu = 285 \). Similarly, part (b) involves a sample mean of 12.025 which is tested against \( \mu = 10.70 \). The exercise uses the sample mean to calculate the test statistic which evaluates the plausibility of the null hypothesis.

Understanding the sample mean provides insight into the general trend of your data and it forms the base for further statistical analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.

In Las Vegas and Atlantic City, New Jersey, tests are performed often on the various gaming devices used in casinos. For example, dice are often tested to determine if they are balanced. Suppose you are assigned the task of testing a die, using a two-tailed test to make sure that the probability of a 2 -spot is \(1 / 6 .\) Using the \(5 \%\) significance level, determine how many 2 -spots you would have to obtain to reject the null hypothesis when your sample size is \(\begin{array}{lll}\text { a. } 120 & \text { b. } 1200 & \text { c. } 12,000\end{array}\) Calculate the value of \(\hat{p}\) for each of these three cases. What can you say about the relationship between (1) the difference between \(\hat{p}\) and \(1 / 6\) that is necessary to reject the null hypothesis and (2) the sample size as it gets larger?

A real estate agent claims that the mean living area of all single-family homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? What will your conclusion be if \(\alpha=.01 ?\)

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 \(\alpha=.10\) ?

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.57, \quad H_{1}: p \neq .57, \quad n=800, \quad \hat{p}=.50, \quad \alpha=.05\) b. \(H_{0}: p=.26, \quad H_{1}: p<.26, \quad n=400, \quad \hat{p}=.23, \quad \alpha=.01\) c. \(H_{0}: p=.84, \quad H_{1}: p>.84, \quad n=250, \quad \hat{p}=.85, \quad \alpha=.025\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.