a) H0: The true average percentage of organic matter in such soil is 3%
H1: The true average percentage of organic matter in such soil is not 3%
Let the los be alpha = 5%
R-code:
Test Statistic t = -1.753 and df = 29 ,
P-value = 0.09016
Here P-value > alpha 0.01 so we accept H0
b) Thus we conclude that the true average percentage of organic matter in such soil is 3%
R Code Value P(T < t-stat) pt(t-value, df) P(T > t-stat) 1-pt (t-value, df) Please do 4 of the following 5 questions. Do more for the practice. If you do all of them, 1 EC on Quiz 12. Also,...
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data. 1.15 5.09 0.97 1.59 4.60 0.32 0.55 1.45 0.10 4.47 1.20 3.50 5.02 4.67 5.22 2.69 3.93 3.17 3.03 2.21 0.69 4.47 3.31 1.17 0.79 1.17 1.57 2.62 1.66 2.05 I USE SALT The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.481, 1.614,...
Give me a solution. Thank you 2. A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data (from "Engineering Properties of 110 509 0.97 1.59 460 0.32 0.55- 1.45- 0.14 14.47- 1 20 13.50 $/02 4.67 5.22 2.69 3.98 3.17 3.03 2.21 0.69 447 3.31 Soil," Soil Science, 1998: 93-102) 1.17 76 | 1.1기 1.5기 2.62 | 1.66 1 2.05 The values...
3. A statistical program is recommended A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data, 1.16 5.09 0.97 1.59 4.60 0.32 0.55 1.45 0.10 4.47 1.20 3.50 5.02 4.67 5.22 2.69 3.98 3.17 3.03 2.21 0.69 4.47 3.31 1.17 0.76 1.17 1.57 2.62 1.66 2.05 The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean...
Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...
Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...