/*! 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 20 Determine the number of degrees ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 20

Determine the number of degrees of freedom for the two-sample \(t\) test or CI in each of the following situations: a. \(m=10, n=10, s_{1}=5.0, s_{2}=6.0\) b. \(m=10, n=15, s_{1}=5.0, s_{2}=6.0\) c. \(m=10, n=15, s_{1}=2.0, s_{2}=6.0\) d. \(m=12, n=24, s_{1}=5.0, s_{2}=6.0\)

Short Answer

Expert verified
a. 18, b. 22, c. 13, d. 31

Step by step solution

01

Introduce the Welch-Satterthwaite Formula

The degrees of freedom (df) for a two-sample t-test can be approximated using the Welch-Satterthwaite formula for unequal variances. The formula is:\[\text{df} = \frac{\left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right)^2}{\frac{\left(\frac{s_1^2}{m}\right)^2}{m-1} + \frac{\left(\frac{s_2^2}{n}\right)^2}{n-1}}.\]
02

Calculate the Combined Variance

Substitute the given values into the combined variance part of the formula \( \left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right) \) for each case:
03

Situation a: Calculate Combined Variance

Substitute \(m=10\), \(n=10\), \(s_1=5.0\), and \(s_2=6.0\):\[\frac{5^2}{10} + \frac{6^2}{10} = \frac{25}{10} + \frac{36}{10} = 6.1\]
04

Situation b: Calculate Combined Variance

Substitute \(m=10\), \(n=15\), \(s_1=5.0\), and \(s_2=6.0\):\[\frac{5^2}{10} + \frac{6^2}{15} = \frac{25}{10} + \frac{36}{15} = 4.933\]
05

Situation c: Calculate Combined Variance

Substitute \(m=10\), \(n=15\), \(s_1=2.0\), and \(s_2=6.0\):\[\frac{2^2}{10} + \frac{6^2}{15} = \frac{4}{10} + \frac{36}{15} = 2.933\]
06

Situation d: Calculate Combined Variance

Substitute \(m=12\), \(n=24\), \(s_1=5.0\), and \(s_2=6.0\):\[\frac{5^2}{12} + \frac{6^2}{24} = \frac{25}{12} + \frac{36}{24} = 3.736\]
07

Calculate Degrees of Freedom

Now, compute the degrees of freedom using the formula for each situation based on already computed combined variances.
08

Situation a: Compute Degrees of Freedom

Using the combined variance 6.1:\[\text{df} = \frac{(6.1)^2}{\frac{(\frac{25}{10})^2}{9} + \frac{(\frac{36}{10})^2}{9}} = 17.51 \approx 18\]
09

Situation b: Compute Degrees of Freedom

Using the combined variance 4.933:\[\text{df} = \frac{(4.933)^2}{\frac{(\frac{25}{10})^2}{9} + \frac{(\frac{36}{15})^2}{14}} = 22.10 \approx 22\]
10

Situation c: Compute Degrees of Freedom

Using the combined variance 2.933:\[\text{df} = \frac{(2.933)^2}{\frac{(\frac{4}{10})^2}{9} + \frac{(\frac{36}{15})^2}{14}} = 12.96 \approx 13\]
11

Situation d: Compute Degrees of Freedom

Using the combined variance 3.736:\[\text{df} = \frac{(3.736)^2}{\frac{(\frac{25}{12})^2}{11} + \frac{(\frac{36}{24})^2}{23}} = 31.33 \approx 31\]

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 91影视!

Key Concepts

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

Two-sample t-test
The two-sample t-test is a statistical method used to determine if there are significant differences between the means of two independent groups. This test is essential when you have two separate samples and want to compare their means to see if they significantly differ from each other. There are two versions of the two-sample t-test:
  • Pooled variance t-test: Assumes that the two groups have equal variances.
  • Welch's t-test: Adjusts for unequal variances and is more reliable when variances between the groups are not equal.
This method is widely applied in various fields, including medicine, social sciences, and more, to test hypotheses about group differences.
Welch-Satterthwaite formula
The Welch-Satterthwaite formula is an adjustment used in the two-sample t-test for samples that have unequal variances. This formula is crucial for calculating the degrees of freedom for the t-test when variances are not assumed to be equal. The formula is expressed as:\[\text{df} = \frac{\left(\frac{s_1^2}{m} + \frac{s_2^2}{n}\right)^2}{\frac{\left(\frac{s_1^2}{m}\right)^2}{m-1} + \frac{\left(\frac{s_2^2}{n}\right)^2}{n-1}}.\]Where:
  • \(s_1\) and \(s_2\): Standard deviations of the two groups.
  • \(m\) and \(n\): Sample sizes of the first and second group, respectively.
This formula helps provide a more flexible and accurate result when handling groups with different variances, thus improving the reliability of statistical inferences.
Statistical inference
Statistical inference refers to the process of using data from a sample to make conclusions about a larger population. It helps researchers make predictions and decisions based on the data they collect. In the context of a two-sample t-test, statistical inference might involve:
  • Determining if the difference between sample means is significant.
  • Estimating the population parameters like the mean or variance.
  • Making data-driven decisions, such as recommending a change in process or treatment based on sample data.
Statistical inference can bridge the gap between data and real-world application, enabling actionable insights despite the variability and uncertainty inherent in sample data.
Unequal variances
Unequal variances, also known as heteroscedasticity, occur when two or more groups being compared have different levels of variability or spread in their data. This can complicate statistical analysis as many traditional tests, like the pooled t-test, assume equal variances. When facing unequal variances, analysts usually:
  • Use statistical tests adjusted for unequal variances, such as Welch's t-test.
  • Apply the Welch-Satterthwaite formula to adjust degrees of freedom in analysis.
Ignoring the presence of unequal variances can lead to incorrect conclusions. It鈥檚 critical in research to identify and properly account for variance differences to ensure accurate results.

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

Researchers sent 5000 resumes in response to job ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2500 of them had "white sounding" first names, such as Brett and Emily, whereas the other 2500 had "black sounding" names such as Tamika and Rasheed. The resumes of the first type elicited 250 responses and the resumes of the second type only 167 responses (these numbers are very consistent with information that appeared in a January 15 , 2003 , report by the Associated Press). Does this data strongly suggest that a resume with a "black" name is less likely to result in a response than is a resume with a "white" name?

Information about hand posture and forces generated by the fingers during manipulation of various daily objects is needed for designing hightech hand prosthetic devices. The article "Grip Posture and Forces During Holding Cylindrical Objects with Circular Grips" (Ergonomics, 1996: 1163-1176) reported that for a sample of 11 females, the sample mean four-finger pinch strength (N) was \(98.1\) and the sample standard deviation was 14.2. For a sample of 15 males, the sample mean and sample standard deviation were \(129.2\) and \(39.1\), respectively. a. A test carried out to see whether true average strengths for the two genders were different resulted in \(t=2.51\) and \(P\)-value \(=.019\). Does the appropriate test procedure described in this chapter yield this value of \(t\) and the stated \(P\)-value? b. Is there substantial evidence for concluding that true average strength for males exceeds that for females by more than \(25 \mathrm{~N}\) ? State and test the relevant hypotheses.

The following data refers to airborne bacteria count (number of colonies/ft \({ }^{3}\) ) both for \(m=8\) carpeted hospital rooms and for \(n=8\) uncarpeted rooms ("Microbial Air Sampling in a Carpeted Hospital," J. Environ. Health, 1968: 405). Does there appear to be a difference in true average bacteria count between carpeted and uncarpeted rooms? $$ \begin{array}{llllllllll} \text { Carpeted } & 11.8 & 8.2 & 7.1 & 13.0 & 10.8 & 10.1 & 14.6 & 14.0 \\ \text { Uncarpeted } & 12.1 & 8.3 & 3.8 & 7.2 & 12.0 & 11.1 & 10.1 & 13.7 \end{array} $$ Suppose you later learned that all carpeted rooms were in a veterans' hospital, whereas all uncarpeted rooms were in a children's hospital. Would you be able to assess the effect of carpeting? Comment.

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar{x}=18.12 \mathrm{kgf} / \mathrm{cm}^{2}\) for the modified mortar \((m=40)\) and \(\bar{y}=16.87 \mathrm{kgf} / \mathrm{cm}^{2}\) for the unmodified mortar ( \(n=32\) ). Let \(\mu_{1}\) and \(\mu_{2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. a. Assuming that \(\sigma_{1}=1.6\) and \(\sigma_{2}=1.4\), test \(H_{0}\) : \(\mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) at level \(.01\). b. Compute the probability of a type II error for the test of part (a) when \(\mu_{1}-\mu_{2}=1\). c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta=.10\) when \(\mu_{1}-\mu_{2}=1\). If \(m=40\), what value of \(n\) is necessary? d. How would the analysis and conclusion of part (a) change if \(\sigma_{1}\) and \(\sigma_{2}\) were unknown but \(s_{1}=1.6\) and \(s_{2}=1.4\) ?

A student project by Heather Kral studied students on "lifestyle floors" of a dormitory in comparison to students on other floors. On a lifestyle floor the students share a common major, and there are a faculty coordinator and resident assistant from that department. Here are the grade point averages of 30 students on lifestyle floors (L) and 30 students on other floors \((\mathrm{N})\) : L: \(2.00,2.25,2.60,2.90,3.00,3.00,3.00,3.00\), \(3.00,3.20,3.20,3.25,3.30,3.30,3.32,3.50\), \(3.50,3.60,3.60,3.70,3.75,3.75,3.79,3.80\), \(3.80,3.90,4.00,4.00,4.00,4.00\). \(\mathrm{N}: 1.20,2.00,2.29,2.45,2.50,2.50,2.50,2.50\), \(2.65,2.70,2.75,2.75,2.79,2.80,2.80,2.80\), \(2.86,2.90,3.00,3.07,3.10,3.25,3.50,3.54\), \(3.56,3.60,3.70,3.75,3.80,4.00\). Notice that the lifestyle grade point averages have a large number of repeats and the distribution is skewed, so there is some question about normality. a. Obtain a \(95 \%\) confidence interval for the difference of population means using the method based on the theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? f. Interpret your results. Is there a substantial difference between lifestyle and other floors? Why do you think the difference is as big as it is?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation
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.