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Chapter 9: Problem 54

Consider \(H_{0}: \mu=80\) versus \(H_{1}: \mu \neq 80\) for a population that is normally distributed. a. A random sample of 25 observations taken from this population produced a sample mean of 77 and a standard deviation of 8 . Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 25 observations taken from the same population produced a sample mean of 86 and a standard deviation of \(6 .\) Using \(\alpha=.01\), would you reject the null hypothesis?

Short Answer

Expert verified
For the first sample, the null hypothesis cannot be rejected. For the second sample, the null hypothesis is rejected.

Step by step solution

01

Compute the test statistic for the first sample

First, apply the formula to calculate the test statistic for the first sample: \(t = (77 - 80) / (8 / \sqrt{25}) = -1.875\).
02

Compare the absolute value of calculated test statistic with the critical value for the first sample

Next, determine the critical value from t-distribution table, which, for \(\alpha=0.01\), and degrees of freedom \(df=25-1=24\), is 2.797. Because \(|-1.875| < 2.797\), don't reject the null hypothesis.
03

Compute the test statistic for the second sample

Now, compute the test statistic for the second sample: \(t = (86 - 80) / (6 / \sqrt{25}) = 5.0\).
04

Compare the absolute value of calculated test statistic with the critical value for the second sample

Then, again determine the critical value from the t-distribution table. Because \(|5.0| > 2.797\), reject the null hypothesis.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It is particularly useful when the population standard deviation is unknown and the sample size is relatively small (< 30). The test assesses if the means of two groups are statistically different from each other.
  • One-sample t-test: Compares the mean of a single sample to a known value (often the population mean).
  • Independent two-sample t-test: Compares the means of two independent samples.
  • Paired sample t-test: Compares means from the same group at different times.
In the provided exercise, we use a one-sample t-test to assess whether the sample mean significantly deviates from the population mean of 80, under the assumption of normal distribution.
null hypothesis
The null hypothesis (\( H_0 \)) is a statement used in statistical hypothesis testing that proposes no significant effect or difference exists in a population. It serves as the starting assumption for the test.
  • The null hypothesis is often denoted as \( H_0 \)
  • In testing, we attempt to either reject \( H_0 \) or fail to reject \( H_0 \).
For this exercise, \( H_0: \mu = 80 \) suggests that the population mean is 80. Our goal is to gather evidence from our sample data to either support or refute this claim. If the t-test shows a significant result, we may reject the null hypothesis in favor of an alternative hypothesis, \( H_1: \mu eq 80 \), which implies that the mean differs from 80.
critical value
The critical value is a threshold that determines the boundary at which a test statistic is considered significantly different from the expected under the null hypothesis. It is derived from the sampling distribution of the test statistic when the null hypothesis is true.
  • Critical values depend on the chosen significance level \( \alpha \)
  • For a two-tailed test, like in our exercise, critical values define both ends of the distribution range.
In the exercise, we use a significance level of \( \alpha = 0.01 \). For a t-distribution with 24 degrees of freedom, the critical value is approximately 2.797. If our test statistic exceeds this value in absolute terms, we consider the sample results significant enough to reject the null hypothesis.
test statistic
The test statistic is a standardized value calculated from sample data during a hypothesis test. It quantifies the degree to which the sample data deviate from the null hypothesis. In a t-test, the test statistic follows a t-distribution, assuming the null hypothesis is true.
  • Calculated using the formula: \( t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \)
  • \( \bar{x} \) is the sample mean, \( \mu \) is the population mean under \( H_0 \), \( s \) is the sample standard deviation, and \( n \) is the sample size.
In the exercise, the test statistic for the first sample is -1.875, and for the second sample, it is 5.0. These values assess how extreme the observed sample means are in comparison to the hypothesized population mean of 80.

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Most popular questions from this chapter

According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

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.

A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is \(49.50\) minutes with a standard deviation of 3 minutes. Assume that the drying times for these paints have a normal distribution. a. Using the \(1 \%\) significance level, would you conclude that the company's claim is true? b. What is the Type I error in this exercise? Explain in words. What is the probability of making such an error?

In a 2009 nonscientific poll on the Web site of the Daily Gazette of Schenectady, New York, readers were asked the following question: "Are you less inclined to buy a General Motors or Chrysler vehicle now that they have filed for bankruptcy?" Of the respondents, \(56.1 \%\) answered "Yes" (http://www. dailygazette.com/polls/2009/jun/Bankruptcy/). In a recent survey of 1200 adult Americans who were asked the same question, 615 answered "Yes." Can you reject the null hypothesis at the \(1 \%\) significance level in favor of the alternative that the percentage of all adult Americans who are less inclined to buy a General Motors or Chrysler vehicle since the companies filed for bankruptcy is different from \(56.1 \%\) ? Use both the \(p\) -value and the critical-value approaches.

Acme Bicycle Company makes derailleurs for mountain bikes. Usually no more than \(4 \%\) of these parts are defective, but occasionally the machines that make them get out of adjustment and the rate of defectives exceeds \(4 \%\). To guard against this, the chief quality control inspector takes a random sample of 130 derailleurs each week and checks each one for defects. If too many of these parts are defective, the machines are shut down and adjusted. To decide how many parts must be defective to shut down the machines, the company's statistician has set up the hypothesis test $$ H_{0}: p \leq .04 \text { versus } H_{1}: p>.04 $$ where \(p\) is the proportion of defectives among all derailleurs being made currently. Rejection of \(H_{0}\) would call for shutting down the machines. For the inspector's convenience, the statistician would like the rejection region to have the form, "Reject \(H_{0}\) if the number of defective parts is \(C\) or more." Find the value of \(C\) that will make the significance level (approximately) \(.05\).
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