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Make the following tests of hypotheses. a. \(H_{0}: \mu=25, \quad H_{1}: \mu \neq 25, \quad n=81, \quad \bar{x}=28.5, \quad \sigma=3, \quad \alpha=.01\) b. \(H_{0}: \mu=12, \quad H_{1}: \mu<12, \quad n=45, \quad \bar{x}=11.25, \quad \sigma=4.5, \quad \alpha=.05\) c. \(H_{0}: \mu=40, \quad H_{1}: \mu>40, \quad n=100, \quad \bar{x}=47, \quad \sigma=7, \quad \alpha=.10\)

Short Answer

Expert verified
In all cases, the test statistic was calculated and then compared to the corresponding critical Z score for the given alpha level. For Hypotheses A and C, the null hypotheses were rejected, while for Hypotheses B, the null hypothesis could not be rejected.

Step by step solution

01

Hypotheses A

First, standardize the test statistic by subtracting the claim \(\mu\) from the sample mean \(\bar{x}\) and dividing by the standard error which is \(\sigma / \sqrt{n}\). The resulting value is \(Z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{28.5 - 25}{3/\sqrt{81}} = 10.5\). Compare it to \(\pm Z_{\alpha/2} = \pm Z_{0.005}\). If it falls in rejection region (\(Z > Z_{0.005}\) or \(Z < -Z_{0.005}\)), we reject \(H_0\). As |10.5| > |2.58| (critical Z score), reject null hypothesis.
02

Hypotheses B

Same procedure: calculate the test statistic \(Z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{11.25 - 12}{4.5/\sqrt{45}} = -1.18\). Now we have a one-tailed test so we will reject \(H_0\) if \(Z < -Z_{\alpha} = -Z_{0.05}\). As -1.18 > -1.645 (critical Z score), fail to reject null hypothesis.
03

Hypotheses C

Once more, standardize: \(Z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{47 - 40}{7/\sqrt{100}} = 10\). Since this is a right-tailed test we'll reject \(H_0\) if \(Z > -Z_{\alpha} = Z_{0.10}\). Given that 10 > 1.28 (critical Z score), reject null hypothesis.

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Key Concepts

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

Test Statistic
When conducting a hypothesis test, the test statistic plays a fundamental role. It's essentially a standardized version of our sample data. By standardizing, we convert the observed data (like sample mean) into a form where we can determine how extreme the observed data is under the null hypothesis.
To calculate the test statistic, we subtract the null hypothesis's claimed population mean \( \mu \) from the sample mean \( \bar{x} \), then divide by the standard error. The formula is given by:
  • \( Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \)
This formula indicates how many standard errors the sample mean \( \bar{x} \) is away from the claimed population mean in the null hypothesis. A larger test statistic (whether positive or negative) signifies evidence against the null hypothesis, especially when it crosses a certain threshold - the critical value.
Null Hypothesis
The null hypothesis \( H_0 \) is a fundamental part of hypothesis testing. It represents a baseline assumption that there is no effect or no difference.
Essentially, it's a statement that believes the observed data comes from a particular population distribution with a specified parameter (like the population mean). Examples include \( H_0: \mu = 25 \) or \( H_0: \mu = 12 \).
In hypothesis testing, we never "accept" the null hypothesis, but we can "reject" it if the evidence strongly contradicts it. Rejecting \( H_0 \) suggests that the alternative hypothesis \( H_1 \), which claims a specific change or effect, is more plausible.
  • \( H_1: \mu eq 25 \) implies the mean differs from 25.
  • \( H_1: \mu < 12 \) suggests the mean is less than 12.
  • \( H_1: \mu > 40 \) indicates the mean is greater than 40.
Understanding and formulating these hypotheses is crucial as they set the foundation of the testing procedure.
Critical Value
In hypothesis testing, after calculating the test statistic, it is compared to a certain threshold known as the critical value. The critical value helps us decide whether to reject or fail to reject the null hypothesis.
It is dependent on two factors: the significance level \( \alpha \) and the nature of the hypothesis (one-tailed or two-tailed). A significant level \( \alpha \) is the probability of rejecting a true null hypothesis, commonly set at 0.01, 0.05, or 0.10 indicating 1%, 5%, or 10% risk respectively.
  • In a two-tailed test, the critical values are split equally in both directions (positive and negative), whereas, in a one-tailed test, it's only in one direction.
  • The Z-score tables or the statistical software are used to find the critical Z value corresponding to the chosen \( \alpha \).
For example, at \( \alpha = 0.05 \) for a one-tailed test, the critical value is \( Z_{0.05} = 1.645 \). If our test statistic is beyond this critical value, we reject the null hypothesis.
Standard Error
The standard error (SE) is a crucial concept when working with sample data in hypothesis testing. It measures the amount of variability or dispersion of a sample statistic. In simpler terms, it's an estimate of how much a sample mean \( \bar{x} \) will differ from the true population mean \( \mu \).
The formula for the standard error of the mean is:
  • \( SE = \frac{\sigma}{\sqrt{n}} \)
Where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
  • A smaller standard error indicates that the sample mean is a more precise estimate of the population mean.
  • As the sample size increases, the standard error decreases, implying more reliable results.
Thus, understanding standard error assists in assessing the reliability and consistency of your sample data in the context of hypothesis testing.

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Most popular questions from this chapter

A random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

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Two years ago, \(75 \%\) of the customers of a bank said that they were satisfied with the services provided by the bank. The manager of the bank wants to know if this percentage of satisfied customers has changed since then. She assigns this responsibility to you. Briefly explain how you would conduct such a test.

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 \(\alpha=.10\) ?

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