Question

The weight percent of silicon in six different rock samples, each containing different amounts of silicon, was measured by two different methods. The results are given in the table. Sample Method A Si wt% Method B Si wt% 11.570 11.540 15.380 15.270 17.950 17.800 22.840 22.810 25.110 25.160 6 27.070 27.010 Determine tcalc. tcalc = | 1.64 Determine ttable at the 95% confidence level. A list of t values can be found in the Student's t table. ttable =

Answer 1

Since we have the same sample, we will use a paired t-test.

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	11.570	11.540	0.030000000000001
	15.380	15.270	0.11
	17.950	17.800	0.15
	22.840	22.810	0.030000000000001
	25.110	25.160	-0.050000000000001
	27.070	27.010	0.059999999999999
Average	19.987	19.932	0.055
St. Dev.	6.013	6.04	0.07
n	6	6	6

For the score differences, we have

D=0,055

Test Statistics

The t-statistic is computed as shown in the following formula:

$t_{calc} = \frac{\bar D}{s_D/ \sqrt n} = \frac{ 0.055}{ 0.07/ \sqrt 6} = 1.931$

Critical Value

(Use two-tailed test since not specified)

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=5.

Hence, it is found that the critical value for this two-tailed test is

table = 2.571

t Table cum. prob one-tail two-tails .50 0.50 1.00 0. 75 0.25 0.50 0 .80 0.20 0.40 85 0.15 0.30 90 0.10 0.20 95 0.05 0.10 975 0.025 0.05 99 0.01 0.02 .995 0.005 0.01 .999 9995 0.001 0.0005 0.002 0.001 AWN 0.000 0.000 0.000 .000 0.000 0.000 1.000 0.816 0.765 0.741 0.727 0.718 1.376 1.061 0.978 0.941 0.920 .906 1.963 1.386 1.250 1.190 1.156 1.134 3.078 1.886 1.638 1.533 1.476 1.440 6.314 2.920 2.353 2.132 2.015 1.943 12.71 4.303 3.182 2.776 2.571 2.447 31.82 6.965 4.541 3.747 3.365 3.143 63.66 9.925 5.841 4.604 4.032 3.707 318.31 22.327 10.215 7.173 5.893 5.208 636.62 31.599 12.924 8.610 6.869 5.959 4 0 5 6 0

Please do upvote if you are satisfied! Let me know in the comments if anything is not clear. I will reply ASAP!