A statistical program is recommended An article reported on various chemical pro

Question

A statistical program is recommended An article reported on various chemical properties of natural and artificial soils. Here are observations on x cation exchange capacity (CEC, in meq/100 g) and y specific surface area (SSA, in m2/g) of 20 natural soils. x 66 121 135 101 77 89 63 57 117 118 76 125 75 71 133 104 76 96 58 109 y 174 323 460 288 205 210 295 161 314 265 236 355 240 133 431 306 132 269 158 303 Minitab gave the following output in response to a request for r correlation of x and y = 0.854 Normal probabilty plots of x and y are quite straight (a) Carry out a test of hypotheses to see if there is a positive linear association in the population from which the sample data was selected State the appropriate null and alternative hypotheses Calculate the test statistic and detenmine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) D-value = State the conclusion in the problem context using 0.01 significance level. Reject Ho. A statistically significant positive correlation does not exist between cation exchange capacity and specific surface area for this class of soils. Reject Ho. A statistically significant positive correlation exists between cation exchange capacity and specific surface area for this class of soils. Fall to reject Ho. A statistically significant positive correlation exists between cation exchange capacity and specific surface area for this class of soils. Fail to reject HO. A statistically significant positive correlation does not exist between cation exchange capacity and specific surface area for this class of soils. (b) with n = 20, how small would the value of r have to be in order for the null hypothesis in the test of (a) to not be rejected at significance level 0.01? (Round your answer to three decimal places.) (c) Calculate a confidence interval for using a 95% confidence level. (Round your answers to three decimal places.)

Explanation / Answer

a) Correct Answer:Option (B) H0: row = 0 Ha : row > 0

Test statistic t = 0.854 / sqrt((1-0.854^2)/(20-2)) = 0.854/0.12263 = 6.9641

P-value = 0.0000
Correct answer: Option (B)

b) t < 0.515

c)
The 95% confidence interval of the population correlation coefficient is
r +/- tcrit SD = (0.854 - 2.101*0.12263, 0.854 + 2.101*0.12263) = (0.59635,1.11165)

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