Question

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 are 2.482, 1.616, and 0.295, respectively. Does this data suggest that the true average percentage of organic matter in such soll is something other than 37 Carry out a test of the appropriate hypotheses at significance level 0.10. Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size] State the appropriate hypotheses. O HOM H. <3 Б) Calculate the test statistic and determine the p-value. (Round your test statistic to two decimal places and your p-value to three decimal places.) P-value c) What can you conclude? Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. O Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. O Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. Would your conclusion be different if a = 0.05 had been used? Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%. O Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage of organic matter in this type of soil is something other than 3%.

Answer 1

c)

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :u =3 against alternative hypothesis H1 :p #3 Let I denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, S =2.4820, 82 = 2.6115, 8 = 1.6160, n = 30, standard error of mean 1.62 30 0.2950 The test statistic can be written as: n(-3) s which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.1 level of significance if P-value < 0.1 or if tobs > to.05,0-1 = 1.699127 0.4 0.3 - density 0.2 0.1- 0.0 - Now, Value of the test statistic: The value of the test statistic tobs 30 (2.4820 - 3) 1.6160 -1.755695 ~ -1.76 associated degrees of freedom = 29 The critical value ==t0,0,n–1 = $t0,050,29 = +1.699127 +1.6991 P - value = P(|tn-11 > tobs) = 2 * P(+29 < -1.755695) = 2*0.044850 = 0.089699 -0.0897 Conclusion : Since p-value < 0.1 and tobs > tcritical = 1.699127, so we reject Ho at 0.1 level of significance Hence, we can conclude that the the population mean is significantly different from 3.

ans-> i)

d)

Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > to.025,n–1 = 2.045230 0.4 0.3 0.2 - 0.1 0.0 - Now, Value of the test statistic: The value of the test statistic tobs = 30 (2.4820 - 3) 1.6160 = -1.755695 -1.76 associated degrees of freedom = 29 The critical value = Et 0.05 = Et0.025,29 = 22.045230 +2.0452 2,0-1 P-value = P(|tn-1| > tobs) = 2 * P(+29 < -1.755695) = 2 * 0.044850 = 0.089699 0.0897 Conclusion : Since p-value > 0.05 and tobs tcritical = 2.045230, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that the population mean is not significantly different from 3.

ans-> iv)