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Chapter 8: Problem 6

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=342\) vs. \(H a: \mu<342, \sigma=11.2, n=40, x-=339, s=10.3\) b. Testing \(H_{0}: \mu=105\) vs. Ha: \(\mu>105, \sigma=5.3, n=80, x-=107, s=5.1\) c. Testing \(H_{0}: \mu=-13.5\) vs. Ha: \(\mu \neq-13.5, \sigma\) unknown, \(n=32, x-=-13.8, s=1.5\) d. Testing \(H_{0}: \mu=28\) vs. \(H a: \mu \neq 28, \sigma\) unknown, \(n=68, x-27.8, s=1.3\)

Short Answer

Expert verified
Test statistics: (a) -1.69, (b) 3.38, (c) -1.13, (d) -1.27.

Step by step solution

01

Understand the Problem and Gather Information for Part a

For part (a), we are conducting a hypothesis test for the population mean \( \mu \) with the null hypothesis \( H_0: \mu = 342 \) and the alternative hypothesis \( H_a: \mu < 342 \). The population standard deviation \( \sigma = 11.2 \), the sample size \( n = 40 \), the sample mean \( \bar{x} = 339 \), and the sample standard deviation \( s = 10.3 \). Since \( \sigma \) is known, a Z-test is appropriate.
02

Compute the Test Statistic for Part a

Use the formula for the Z-test statistic for the mean: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Substitute the given values: \( \bar{x} = 339 \), \( \mu = 342 \), \( \sigma = 11.2 \), \( n = 40 \). \[ Z = \frac{339 - 342}{\frac{11.2}{\sqrt{40}}} = \frac{-3}{1.77} \approx -1.69 \].
03

Gather Information for Part b

For part (b), we have the hypotheses \( H_0: \mu = 105 \) and \( H_a: \mu > 105 \). The population standard deviation \( \sigma = 5.3 \), the sample size \( n = 80 \), the sample mean \( \bar{x} = 107 \), and the sample standard deviation \( s = 5.1 \). As \( \sigma \) is known, a Z-test is used.
04

Compute the Test Statistic for Part b

We use the same Z-test statistic formula from Step 2: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Plug in the numbers: \( \bar{x} = 107 \), \( \mu = 105 \), \( \sigma = 5.3 \), \( n = 80 \). \[ Z = \frac{107 - 105}{\frac{5.3}{\sqrt{80}}} = \frac{2}{0.5921} \approx 3.38 \].
05

Gather Information for Part c

For part (c), the hypotheses are \( H_0: \mu = -13.5 \) and \( H_a: \mu eq -13.5 \). The population standard deviation is unknown, the sample size \( n = 32 \), the sample mean \( \bar{x} = -13.8 \), and the sample standard deviation \( s = 1.5 \). As \( \sigma \) is unknown, a T-test is applicable.
06

Compute the Test Statistic for Part c

Use the T-test statistic formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Substitute the values: \( \bar{x} = -13.8 \), \( \mu = -13.5 \), \( s = 1.5 \), \( n = 32 \). \[ t = \frac{-13.8 + 13.5}{\frac{1.5}{\sqrt{32}}} = \frac{-0.3}{0.265} \approx -1.13 \].
07

Gather Information for Part d

For part (d), the hypotheses are \( H_0: \mu = 28 \) and \( H_a: \mu eq 28 \). The population standard deviation is unknown, the sample size \( n = 68 \), the sample mean \( \bar{x} = 27.8 \), and the sample standard deviation \( s = 1.3 \). A T-test is required because \( \sigma \) is unknown.
08

Compute the Test Statistic for Part d

Apply the T-test statistic formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Insert the numbers: \( \bar{x} = 27.8 \), \( \mu = 28 \), \( s = 1.3 \), \( n = 68 \). \[ t = \frac{27.8 - 28}{\frac{1.3}{\sqrt{68}}} = \frac{-0.2}{0.15764} \approx -1.27 \].

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Key Concepts

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

Z-test
When we talk about the Z-test, we are diving into a statistical method used to determine if there is a significant difference between sample and population means. This is particularly useful when the population standard deviation is known and the sample size is large, usually considered as 30 or more. The Z-test is grounded in the concept of the normal distribution and uses the Z-score formula for its calculations.

Here is how it works:
  • Formula: The Z-test statistic is calculated using the formula: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where:
    • \( \bar{x} \) is the sample mean,
    • \( \mu \) is the population mean,
    • \( \sigma \) is the population standard deviation, and
    • \( n \) is the sample size.
  • Usage: The Z-test is typically used for testing hypotheses such as:
    • Whether a sample mean is different from a known or hypothesized population mean.
    • Comparing means from two different samples assuming common variance.
  • Interpretation: The result, known as the Z-score, shows how many standard deviations the sample mean is from the population mean.
    • A Z-score near 0 indicates the sample mean is similar to the population mean.
    • A larger positive or negative Z-score indicates a significant difference.
By calculating the Z-score, we can determine the probability of observing the test results under the null hypothesis, which helps in making an informed decision about the hypotheses.
T-test
The T-test is a paramount statistical tool used when we do not know the population standard deviation and must rely on the sample standard deviation. It's an ideal choice for smaller sample sizes, generally considered to be less than 30, and when the population standard deviation is unknown.

This method aims to determine whether there is a significant difference between the means of two groups or conditions.
  • Formula: To calculate the T-test statistic, we use:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]where:
    • \( \bar{x} \) is the sample mean,
    • \( \mu \) is the hypothesized population mean,
    • \( s \) is the sample standard deviation, and
    • \( n \) is the sample size.
  • Application: The T-test is applicable in scenarios such as:
    • Comparing the sample mean to the population mean when standard deviation is unknown.
    • Comparing means from two related groups.
  • Types: There are different types of T-tests for various conditions:
    • One-sample T-test: compares the sample mean against a known mean.
    • Independent two-sample T-test: compares means from two independent groups.
    • Paire d T-test: focuses on the means from the same group at different times.
The critical aspect of the T-test is the use of the T-distribution, which, unlike the normal distribution, adjusts for sample size, making it a powerful tool for smaller data sets.
Test Statistic
The test statistic is a crucial figure in hypothesis testing. It defines the degree to which your sample data deviates from the null hypothesis. Used in both the Z-test and the T-test, the test statistic helps determine the probability of observing your test results under the null hypothesis.

Let's break it down:
  • Purpose: The test statistic acts as a bridge from the data to the decision.
    • It simplifies the data into a single, interpretable number.
    • Guides you in accepting or rejecting the null hypothesis based on comparison with critical values.
  • Calculation: Each type of test has its own formula for test statistics:
    • Z-test: Dependent on the standard normal distribution, calculates the Z-score.
    • T-test: Based on the T-distribution, calculates the t-value which adjusts for sample size.
  • Decision-making:
    • The magnitude of the test statistic is compared against a critical value from statistical tables.
    • If it exceeds the critical value, the null hypothesis is often rejected.
Overall, the test statistic is essential in quantifying how likely your sample results would occur under the null hypothesis. It translates data observations into actionable decisions in statistical hypothesis testing.

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Most popular questions from this chapter

In the past, \(68 \%\) of a garage's business was with former patrons. The owner of the garage samples 200 repair invoices and finds that for only 114 of them the patron was a repeat customer. a. Test whether the true proportion of all current business that is with repeat customers is less than \(68 \%,\) at the \(1 \%\) level of significance. b. Compute the observed significance of the test.

Large Data Set 6 records results of a random survey of 200 voters in each of two regions, in which they were asked to express whether they prefer Candidate \(A\) for a U.S. Senate seat or prefer some other candidate. Use the full data set (400 observations) to test the hypothesis that the proportion \(p\) of all voters who prefer Candidate \(A\) exceeds \(0.35 .\) Test at the \(10 \%\) level of significance.

An automobile manufacturer recommends oil change intervals of 3,000 miles. To compare actual intervals to the recommendation, the company randomly samples records of 50 oil changes at service facilities and obtains sample mean 3,752 miles with sample standard deviation 638 miles. Determine whether the data provide sufficient evidence, at the \(5 \%\) level of significance, that the population mean interval between oil changes exceeds 3,000 miles.

A magazine publisher tells potential advertisers that the mean household income of its regular readership is $$\$ 61,500.$$ An advertising agency wishes to test this claim against the alternative that the mean is smaller. A sample of 40 randomly selected regular readers yields mean income $$\$ 59,800$$ with standard deviation $$\$ 5,850.$$ Perform the relevant test at the \(1 \%\) level of significance.

The mean score on a 25-point placement exam in mathematics used for the past two years at a large state university is 14.3. The placement coordinator wishes to test whether the mean score on a revised version of the exam differs from 14.3 . She gives the revised exam to 30 entering freshmen early in the summer; the mean score is 14.6 with standard deviation 2.4 a. Perform the test at the \(10 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(10 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).
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