Ch. 9 - Inferences from Two Samples

Triola - Elementary Statistics 14th Edition

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Triola 14th Edition

Ch. 9 - Inferences from Two Samples

Problem 9.2.4

Chapter 9, Problem 9.2.4

Degrees of Freedom For Example 1, we used df=smaller of n1-1 and n2-1 we got df=11 and the corresponding critical value is t=-1.796 (found from Table A-4). If we calculate df using Formula 9-1, we get df=19.370 and the corresponding critical value is t=-1.727 How is using the critical value of t=-1.796 “more conservative” than using the critical value of t=-1.727

Verified step by step guidance

Step 1: Understand the concept of degrees of freedom (df). Degrees of freedom represent the number of independent values or quantities that can vary in a statistical calculation. In this problem, df is calculated using two methods: the smaller of n1-1 and n2-1, and Formula 9-1.

Step 2: Recognize the relationship between degrees of freedom and the critical value of t. A smaller df typically results in a larger critical value of t, which corresponds to a wider confidence interval or stricter threshold for rejecting the null hypothesis.

Step 3: Compare the two critical values provided: t = -1.796 (using df = 11) and t = -1.727 (using df = 19.370). The critical value of t = -1.796 is larger in magnitude, meaning it sets a stricter threshold for statistical significance.

Step 4: Understand the term 'more conservative.' In statistics, a 'more conservative' approach means being less likely to reject the null hypothesis. Using the larger critical value (t = -1.796) requires stronger evidence to reject the null hypothesis, making this approach more conservative.

Step 5: Conclude that using the smaller df (df = 11) and the corresponding critical value of t = -1.796 is more conservative because it increases the likelihood of retaining the null hypothesis, reducing the risk of Type I error (false positive).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Degrees of Freedom

Degrees of freedom (df) refer to the number of independent values or quantities that can vary in a statistical calculation. In hypothesis testing, df is crucial for determining the appropriate distribution to use, as it affects the shape of the t-distribution. The smaller the df, the wider the t-distribution, which can lead to more conservative estimates in statistical tests.

Critical Value

A critical value is a threshold that determines the boundary for rejecting the null hypothesis in hypothesis testing. It is derived from the chosen significance level (alpha) and the relevant statistical distribution, such as the t-distribution. In this context, a more conservative critical value means that it is less likely to reject the null hypothesis, thus reducing the risk of Type I errors.

Conservativeness in Statistical Testing

In statistical testing, a conservative approach refers to using stricter criteria for making decisions, such as requiring stronger evidence to reject the null hypothesis. This is often achieved by using a higher critical value, which leads to a lower probability of falsely rejecting the null hypothesis. In the given example, using the critical value of t=-1.796 is more conservative than t=-1.727, as it requires more substantial evidence to conclude a significant effect.

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Step 2: Calculate Test Statistic

Related Practice

Textbook Question

Body Temperatures Listed below are body temperatures from six different subjects measured at two different times in a day (from Data Set 5 “Body Temperatures” in Appendix B).

a. Are the two sets of data independent or dependent? Explain.

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Textbook Question

In Exercises 5–8, use (a) randomization and (b) bootstrapping for the indicated exercise from Section 9-1. Compare the results to those obtained in the original exercise.

Exercise 8 in Section 9-1 “Tennis Challenges”

Textbook Question

Testing Effects of Alcohol Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 22 people who drank ethanol and another group of 22 people given a placebo. The errors for the treatment group have a standard deviation of 2.20, and the errors for the placebo group have a standard deviation of 0.72 (based on data from “Effects of Alcohol Intoxication on Risk Taking, Strategy, and Error Rate in Visuomotor Performance,” by Streufert et al., Journal of Applied Psychology, Vol. 77, No. 4). Use a 0.05 significance level to test the claim that both groups have the same amount of variation among the errors.

Textbook Question

In Exercises 5–8, use (a) randomization and (b) bootstrapping for the indicated exercise from Section 9-1. Compare the results to those obtained in the original exercise.

Exercise 9 in Section 9-1 “Cell Phones and Handedness”

Textbook Question

Is Friday the 13th Unlucky? Listed below are numbers of hospital admissions in one region due to traffic accidents on different Fridays falling on the 6th day of a month and the following 13th day of the month (based on data from “Is Friday the 13th Bad for Your Health,” by Scanlon et al., British Medical Journal, Vol. 307). Assume that we want to use a 0.05 significance level to test the claim that the data support the claim that fewer hospital admissions due to traffic accidents occur on Friday the 6th than on the following Friday the 13th. Identify the null hypothesis and alternative hypothesis.

Textbook Question

Degrees of Freedom In Exercise 20 “Blanking Out on Tests,” using the “smaller of n1-1 and n2-1” for the number of degrees of freedom results in df=15 Find the number of degrees of freedom using Formula 9-1. In general, how are hypothesis tests and confidence intervals affected by using Formula 9-1 instead of the “smaller of n1-1 and n2-1 ”?