Individual # 3 Solution

Individual # 3 Solution - Solution to Individual Assignment...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution to Individual Assignment # 3 Q6.126 a. Some preliminary calculations: x = 79.93/5 = 15.986 s 2 = 1,277.7627 - 79.93² / 5 5 –1 = .00043 s = √.00043 = .0207 To determine if the mean measurement differs from 16.01, we test: H 0: μ = 16.01 H a: μ ≠ 16.01 The test statistic is t = 15,986 - 16.01 0.0207 /√ 5 = 2.59 The rejection region requires α /2 = .02/2 = .01 in each tail of the t -distribution with df = n 1 = 5 1 = 4. From Table VI, Appendix B, ± t .01;4 =± 3.747. The rejection region is t < 3.747 or t > 3.747. Since the observed value of the test statistic does not fall in the rejection region ( t = 2.59> 3.747 ), H 0 is not rejected. There is insufficient evidence to indicate the true mean measurement differs from 16.01 at α = .02. b. We must assume that the sample of measurements was randomly selected from a population of measurements that is normally distributed.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
c. 98 % C I for mean is
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/23/2011 for the course COMM 215 taught by Professor Ghatri during the Spring '08 term at Concordia Canada.

Page1 / 4

Individual # 3 Solution - Solution to Individual Assignment...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online