{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Individual # 3 Solution

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

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

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.

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

View Full Document
c. 98 % C I for mean is 15.986 ± 3.747 * 0.0207/√ 5 = 15.986 ± 0.0346 = ( 15.952 , 16.0206 ) d. Since (15.952 , 16.0206 ) includes 16.01 therefore we Do Not Reject H 0. at α = .02. e. p- value = 2P(t> | -2.59 | ) 0.025 < area < 0.05 => 0.02 <0.05 < p-value< 0.10 => 0.02<p-value Since the p- value > α = .02 Do Not Reject H 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}