Question

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Air pollution concentrations at a location were measured. Eleven measurements were taken using two different methods and the following values in parts per million of CO2 were obtained: Method A: Method B: 66.3 71.3 63.5 60.4 64.9 64.6 61.8 63.9 64.3 68.8 64.7 70.1 65.1 64.8 64.5 68.9 68.4 65.8 63.2 66.2 67.4 69.2 It is desired to show that the pollution values obtained using Method A are not significantly different from that using Method B at the 5% significance level. The following three tests were carried out in using a computer. Test I: Paired T for Method A - Method B Mean St Dev Method A 11 64.918 1.885 Method B 11 66.727 3.238 Difference -1.809 3.016 (Paired test) SE Mean 0.568 0.976 0 .909 11 T-Test of mean difference - 0 (vs < 0): P-value = 0.037 Test 2: TWO-Sample T for Method A vs Method B (One-tailed test) N Mean St Dev SE Mean Method A 11 64.92 1.88 0.57 Method B 11 66.73 0.98 Difference = mu Method A - mu Method B Estimate for difference: -1.81 T-Test of difference - 0 (vs <): P-value = 0.064 DF - 16 N Test 3: Two-sample T for Method A vs Method 3 (Two-tailed test) Mean St Dev SE Mean Method A 11 64.92 1.88 0.57 Method B 11 66.73 3.24 0.98 Difference = mu Method A - mu Method B Estimate for difference: -1.81 T-Test of difference = 0 (vs not =): P-value = 0.129 DF - 16 Which of the above results is the appropriate one? Explain why this is so. What is the conclusion from the test? What is the main assumption of the t-test used? Use a simple graphical method to verify the assumption of the t-test used? Calculate the value of the test statistic for each of the above tests.

Answer 1

a) Test 3 is the appropriate one. Note we want to test if Two methods are DIFFERENT or not; so one-tailed i.e. Test 2 is ruled out. Also, each of our sample are independent, so paired test i.e. Test 1 is also ruled out.

Conclusion : Since, p-value = 0.129 > significance level = 0.05 ; We ACCEPT the null hypothesis i.e. Two Methods are NOT DIFFERENT or the population mean of measurements from each method are same.

b) Main Assumption : Data from each sample is normally distributed. We verify this using Normal QQ plot as shown below. Both plots give approximately 45 degree line, so assumption is verified.

Method A QQ plot :

Normal Q-Q Plot 68 66 Sample Quantiles 64 62 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles

Method B QQ plot :

Normal Q-Q Plot 68 Sample Quantiles TTTTT 64 60 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles

C) For Test 1 :

We have to find Difference vector : d = XA - XB ,

i	d=Xa - Xb
1	-5
2	3.1
3	0.3
4	-2.1
5	-4.5
6	-5.4
7	0.3
8	-4.4
9	2.6
10	-3
11	-1.8

d = -1.809; sdd = 3.0161; sdā = 3.0161/V11 = 0.90938 test stat ttest = = -1.809/0.90938 = -1.989 sd

Test 2 :

ta - rb SE SE = Vsa/na + sb2/nb = 1.1294; 64.92 - 66.73 ttest = 1.1294 -= -1.6026

Test 3 :

ta - rb SE SE = Vsa/na + sb2/nb = 1.1294; 64.92 - 66.73 ttest = 1.1294 -= -1.6026

Test 2 and 3 will have same test statistic.