EFSC STA 2023 MML ACCESS
EFSC STA 2023 MML ACCESS
6th Edition
ISBN: 9780135927885
Author: Triola
Publisher: PEARSON
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Chapter 11.1, Problem 25BB

Chapter 11.1, Problem 25BB, Assumed mid-point x=fxn=39825180=221.25 s=[fx2n1(fx)2n(n1)]=[10244375179(39825)2180179]=89.48 25. , example  1

* Assumed mid-point

x ¯ = Σ f x n = 39   825 180 = £ 221.25

s = [ Σ f x 2 n 1 ( Σ f x ) 2 n ( n 1 ) ] = [ 10  244375 179 ( 39  825 ) 2 180 × 179 ] = £ 89.48

25. Testing Goodness-of-Fit with a Normal Distribution Refer to Data Set 1 “Body Data” in Appendix B for the heights of females.

Chapter 11.1, Problem 25BB, Assumed mid-point x=fxn=39825180=221.25 s=[fx2n1(fx)2n(n1)]=[10244375179(39825)2180179]=89.48 25. , example  2

a. Enter the observed frequencies in the table above.

b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, use the methods of Chapter 6 to find the probability of a randomly selected height belonging to each class.

c. Using the probabilities found in part (b), find the expected frequency for each category.

d. Use a 0.01 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population?

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Chapter 11 Solutions

EFSC STA 2023 MML ACCESS

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