Explanation Given info: The data represents the random sample of 22 bottles of juice measured the content of bottles. The sample size is n = 22 , the mean amount of juice is x ¯ = 64.01 , standard deviation s = 0.045 and the level of significance are α = 0.01 . Calculation: Requirement check: The sample values obtained from the bottles is assumed to be independent and the sample taken from the machine is a simple random sample. The sample is assumed to be distributed normally. Thus, the condition is satisfied. The test hypotheses are given below: Null hypothesis: H 0 : μ = 64.05 Therefore, the mean amount of juice is 64.05. Alternative hypothesis: H 1 : μ ≠ 64.05 Therefore, the mean amount of juice is different from 64.05

Fundamentals of Statistics (5th Edition)

5th Edition

ISBN: 9780134508306

Author: Michael Sullivan III

Publisher: PEARSON

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Chapter 10.3, Problem 28AYU

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To obtain: The 95% confidence interval about the mean amount of juice.

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Chapter 10.3, Problem 27AYU

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Chapter 10 Solutions

Fundamentals of Statistics (5th Edition)

Ch. 10.1 - A _____ is a statement regarding a characteristic...Ch. 10.1 - _____ _____ is a procedure, based on sample...Ch. 10.1 - The _____ _____ is a statement of no change, no...Ch. 10.1 - The _____ _____ is a statement we are trying to...Ch. 10.1 - If we reject the null hypothesis when the...Ch. 10.1 - If we do not reject the null hypothesis when the...Ch. 10.1 - The _____ _____ _____ is the probability of making...Ch. 10.1 - True or False: Sample evidence can prove a null...Ch. 10.1 - In Problems 914, the null and alternative...Ch. 10.1 - In Problems 914, the null and alternative...

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To obtain: The 95% confidence interval about the mean amount of juice.

Solution Summary: The author explains how to obtain the 95% confidence interval for the mean amount of juice using the MINITAB software.

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To obtain: The 95% confidence interval about the mean amount of juice.

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Chapter 10 Solutions