Explanation Given Info: Part a: We are given to use a two-tailed test. That's why alternative hypothesis has ≠ sign. n = 50 − 11 = 39 x = 29 Part b: We are given to use a two-tailed test. That's why alternative hypothesis has ≠ sign. n = 50 x = 29 + 11 2 = 34.5 Formula used: p ^ = x n z = p ^ − p 0 p 0 ( 1 − p 0 ) n Calculation: Part a: STEP 1: State the hypotheses. The null hypothesis states that these data indicate no significant difference from what would be expected by chance. In symbols: H 0 : p = 0.5 The alternative hypothesis states that these data indicate significant difference from what would be expected by chance. In symbols: H a : p ≠ 0.5 STEP 2: Locate the critical region. From z table, the critical value of z values for the critical region are z = ± 1.96 at α = 0.05 . STEP 3: Compute the test statistic. The sample proportion is: p ^ = x n = 29 39 = 0.74 The test statistic is: z = p ^ − p 0 p 0 ( 1 − p 0 ) n = 0

Statistics for The Behavioral Sciences (MindTap Course List)

10th Edition

ISBN: 9781305504912

Author: Frederick J Gravetter, Larry B. Wallnau

Publisher: Cengage Learning

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Types of Bias

Statistics is the study of data collection, organization, analysis, interpretation, and presentation. Statistical bias is a characteristic of a statistical technique or its findings in which the expected value deviates from the actual root quantitative p…

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Chapter 18, Problem 25P

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Do these data indicate a significant effect of frequency and severity when participants showing no change are discarded?
Do these data indicate a significant effect of frequency and severity when participants showing no change are split between the two groups?

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Chapter 18 Solutions

Statistics for The Behavioral Sciences (MindTap Course List)

Ch. 18 - 1. Insurance companies charge young drivers more...Ch. 18 - 2. Güven, Elaimis, Binokay, and Tan (2003) studied...Ch. 18 - In problem 9 in Chapter 17, we described a study...Ch. 18 - 4. Problems 5 and 6 in Chapter 17 cited a study...Ch. 18 - 5. A researcher would like to determine whether...Ch. 18 - 6. A recent survey of practicing psychotherapists...Ch. 18 - 7. In 2005, Fung et al. published a study...Ch. 18 - 8. In problem 13 in Chapter 9, wepresented a study...Ch. 18 - Prob. 9P Ch. 18 - Prob. 10P

Ch. 18 - Prob. 11P Ch. 18 - 12. One of the original methods for testing ESP...Ch. 18 - Prob. 13P Ch. 18 - Prob. 14P Ch. 18 - 15. In Example 11.2 (p. 343) we presented a...Ch. 18 - Stressful or traumatic experiences can often...Ch. 18 - Prob. 17P Ch. 18 - Prob. 18P Ch. 18 - Prob. 19P Ch. 18 - In Problem 14 in Chapter 11, we described a study...Ch. 18 - In problem 13 in Chapter 17, we Discussed a study...Ch. 18 - Prob. 22P Ch. 18 - Prob. 23P Ch. 18 - Prob. 24P Ch. 18 - Prob. 25P

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Do these data indicate a significant effect of frequency and severity when participants showing no change are discarded? Do these data indicate a significant effect of frequency and severity when participants showing no change are split between the two groups?

Solution Summary: The author explains how the data indicate a significant effect of frequency and severity when participants showing no change are split between the two groups.

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Do these data indicate a significant effect of frequency and severity when participants showing no change are discarded? Do these data indicate a significant effect of frequency and severity when participants showing no change are split between the two groups?

Chapter 18 Solutions

Do these data indicate a significant effect of frequency and severity when participants showing no change are discarded?
Do these data indicate a significant effect of frequency and severity when participants showing no change are split between the two groups?