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Chapter 11: Problem 2

Perform the appropriate hypothesis test. If \(n_{1}=31, s_{1}=12, n_{2}=51,\) and \(s_{2}=10,\) test whether \(\sigma_{1}>\sigma_{2}\) at the \(\alpha=0.05\) level of significance.

Short Answer

Expert verified
Fail to reject H_0. There is not enough evidence to conclude that 蟽_1 > 蟽_2 at 伪 = 0.05.

Step by step solution

01

State the hypotheses

We need to test whether the population standard deviations are different. The null hypothesis (H_0) and the alternative hypothesis (H_A) are as follows:H_0: 蟽_1 鈮� 蟽_2H_A: 蟽_1 > 蟽_2.
02

Determine the appropriate test statistic

Since we are comparing two variances, we will use the F-test for variances. The test statistic is given by:F = (s_1^2 / s_2^2).
03

Calculate the test statistic

Using the given sample standard deviations, we find the test statistic as:F = (12^2 / 10^2) = (144 / 100) = 1.44.
04

Determine the degrees of freedom

The degrees of freedom for the numerator is (n_1 - 1) and for the denominator is (n_2 - 1). Thus we have:df1 = 31 - 1 = 30df2 = 51 - 1 = 50.
05

Find the critical value

Using the F-distribution table at a significance level of 伪 = 0.05 with df1 = 30 and df2 = 50, the critical value is approximately 1.72.
06

Compare the test statistic to the critical value

Since the calculated test statistic F = 1.44 is less than the critical value F_{0.05,30,50} = 1.72, we fail to reject the null hypothesis.
07

State the conclusion

At the 伪 = 0.05 level of significance, we do not have sufficient evidence to conclude that 蟽_1 is greater than 蟽_2.

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Key Concepts

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

F-test
The F-test is a statistical method used to compare two variances and determine if they are significantly different from each other. In hypothesis testing, particularly in variance comparison, the F-test helps determine if the variability within one sample is substantially different from another. The test is named after Sir Ronald Fisher, who developed it.

To perform an F-test, you first calculate the test statistic, denoted as F. The formula for the F-statistic when comparing two samples is: \F = \frac{s1^2}{s2^2}\. Here, \(s1\) and \(s2\) are the sample standard deviations of the first and second groups, respectively. Squaring these values yields the variances.

Once you have the F-value, you compare it to a critical value from the F-distribution table that corresponds to your chosen level of significance and degrees of freedom for each sample.
Variances Comparison
Comparing variances is crucial in statistics, especially when determining if two populations differ in dispersion. Variance measures the degree to which each number in a dataset differs from the mean. When you have samples from two populations and want to compare their variances, the F-test is the appropriate method.

In the scenario given, where \(n_1=31\), \(s_1=12\), \(n_2=51\), and \(s_2=10\), the variances are calculated by squaring the standard deviations: \(s1^2 = 144\) and \(s2^2 = 100\). The F-statistic, therefore, is the ratio of these variances: \(F = \frac{144}{100} = 1.44\).

The next step in comparing variances is to determine the degrees of freedom for each sample. Degrees of freedom refer to the number of values that are free to vary in the calculation of a statistic. In variance comparison, the degrees of freedom for the numerator (\(df1\)) is \(n_1 - 1\), and for the denominator (\(df2\)) is \(n_2 - 1\). For our example, \(df1 = 30\) and \(df2 = 50\).
Critical Value Determination
In hypothesis testing, determining the critical value is vital for comparing it with your test statistic. The critical value is derived from the F-distribution table and hinges on the chosen level of significance (\(\alpha\)) and the degrees of freedom for both the numerator and the denominator.

For our problem, the significance level \(\alpha\) is 0.05, with \(df1 = 30\) and \(df2 = 50\). Using these parameters, you find the critical value from the F-distribution table. It is approximately 1.72.

With the critical value known, compare it to your computed F-statistic (here, 1.44). If the F-statistic exceeds the critical value, you reject the null hypothesis. In our example, since the F-statistic (1.44) is less than the critical value (1.72), we fail to reject the null hypothesis. This means we do not have enough evidence to claim that the variance of the first sample is greater than that of the second one at the 0.05 level of significance.

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Most popular questions from this chapter

Perform the appropriate hypothesis test. If \(n_{1}=61, s_{1}=18.3, n_{2}=57,\) and \(s_{2}=13.5,\) test whether the population standard deviations differ at the \(\alpha=0.05\) level of significance.

Putting It Together: Online Homework Professor Stephen Zuro of Joliet Junior College wanted to determine whether an online homework system (meaning students did homework online and received instant feedback with helpful guidance about their answers) improved scores on a final exam. In the fall semester, he taught a precalculus class using an online homework system. In the spring semester, he taught a precalculus class without the homework system (which meant students were responsible for doing their homework the old- fashioned way - paper and pencil). Professor Zuro made sure to teach the two courses identically (same text, syllabus, tests, meeting time, meeting location, and so on ). The table summarizes the results of the two classes on their final exam $$ \begin{array}{lcc} & \text { Fall Semester } & \text { Spring Semester } \\ \hline \text { Number of students } & 27 & 25 \\ \hline \text { Mean final exam score } & 73.6 & 67.9 \\ \hline \text { Standard deviation } & 10.3 & 12.4 \\ \text { final exam score } & & \end{array} $$ (a) What type of experimental design is this? (b) What is the response variable? What are the treatments in the study? (c) What factors are controlled in the experiment? (d) In many experiments, the researcher will recruit volunteers and randomly assign the individuals to a treatment group. In what regard was this done for this experiment? (e) Did the students perform better on the final exam in the fall semester? Use an \(\alpha=0.05\) level of significance. (f) Can you think of any factors that may confound the results? Could Professor Zuro have done anything about these confounding factors?

Walking in the Airport, Part I Do people walk faster in the airport when they are departing (getting on a plane) or when they are arriving (getting off a plane)? Researcher Seth B. Young measured the walking speed of travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Direction of Travel } & \text { Departure } & \text { Arrival } \\ \hline \text { Mean speed (feet per minute) } & 260 & 269 \\ \hline \begin{array}{l} \text { Standard deviation } \\ \text { (feet per minute) } \end{array} & 53 & 34 \\ \hline \text { Sample size } & 35 & 35 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) Explain why it is reasonable to use Welch's \(t\) -test. (c) Do individuals walk at different speeds depending on whether they are departing or arriving at the \(\alpha=0.05\) level of significance?

Low-birth-weight babies are at increased risk of respiratory infections in the first few months of life and have low liver stores of vitamin A. In a randomized, double-blind experiment, 130 lowbirth-weight babies were randomly divided into two groups. Subjects in group 1 (the treatment group, \(n_{1}=65\) ) were given 25,000 IU of vitamin A on study days \(1,4,\) and 8 where study day 1 was between 36 and 60 hours after delivery. Subjects in group 2 (the control group, \(n_{2}=65\) ) were given a placebo. The treatment group had a mean serum retinol concentration of 45.77 micrograms per deciliter \((\mu \mathrm{g} / \mathrm{dL}),\) with a standard deviation of \(17.07 \mu \mathrm{g} / \mathrm{dL}\). The control group had a mean serum retinol concentration of \(12.88 \mu \mathrm{g} / \mathrm{dL}\), with a standard deviation of \(6.48 \mu \mathrm{g} / \mathrm{dL}\). Does the treatment group have a higher standard deviation for serum retinol concentration than the control group at the \(\alpha=0.01\) level of significance? It is known that serum retinol concentration is normally distributed.

Walking in the Airport, Part II Do business travelers walk at a different pace than leisure travelers? Researcher Seth B. Young measured the walking speed of business and leisure travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Type of Traveler } & \text { Business } & \text { Leisure } \\ \hline \text { Mean speed (feet per minute) } & 272 & 261 \\ \hline \begin{array}{l} \text { Standard deviation (feet per } \\ \text { minute) } \end{array} & 43 & 47 \\ \hline \text { Sample size } & 20 & 20 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) What must be true regarding the populations to use Welch's \(t\) -test to compare the means? (c) Assuming that the requirements listed in part (b) are satisfied, determine whether business travelers walk at a different speed from leisure travelers at the \(\alpha=0.05\) level of significance.
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