# Essay on Science: Statistics and Regression Model

Submitted By SibiGanesh1
Words: 1169
Pages: 5

Question One A)
Exposed:
N=2904 , Bladder Cancer = 40
Πyes= 40/2904
=0.013774104
Not Exposed:
Πno=23/3601
=0.006387114
Point Estimate Calculation:
Pyes – Pno
=0.013774104-0.006387114
=0.00738699
Standard Error Calculation
SE(Pyes-Pno)=√var(Pyes)-Var(Pno)
=√ (Pyes(1-Pyes)/Nyes) + (Pno(1-Pno)/Nno)
=√ (0.013774104(1-0.013774104)/2904) + 0.006387114(1-0.006387114)/3601)
=√ ((0.013774104) +(0.986225896)/2904) + ((0.006387114) (0.99361286)/3601)
=√ ((0.013584378)/2908) + ((0.006346318)/3601)
=√ 0.000004677+ 0.000001762
=√ 0.000006439
SE=0.002537518
Next to calculate Nyes * Pyes ,Nyes (1-Pyes), Nno*Pno, Nno(1-Pno) to check for normality

Nyes * Nno
=2904*0.013774104
=39.99999802

Nyes (1-Pyes)
=2904(1-0.013774104)
=2904(0.986225896)
=2864.000002

Nno * Pno
=3601(0.006387114)
=22.99999751

Nno(1-Pno)
=3601(1-0.006387114)
=3601(0.993612886)
=3578.000002

Since these four values are higher than 5, we can assume normality.
Next we calculate the confidence interval
Α=0.02
We divide 0.02 by 2 so =0.01
CI = (Pyes – Pno) +/- Z0.01 [SE(Pyes-Pno)]

=0.00735699 +/- 2.33 [0.002537518] this value was calculated earlier
=(0.00738699 + 2.33 [0.002537518]) (0.00738699 -2.33[0.002537518]
= (0.00738699 + 0.0059124169), (0.00738699 -0.005912416)
CI=(0.013299406, 0.001474574)
Since 0 is not in our confidence interval we can assume that there is a statistically significant difference between the two groups. B)
Ho : πyes – πno = 0
H1 :πyes – πno ≠ 0 α=0.01 Continuity Correlation:
Ppooled = Ryes + Rno/Nyes + Nno
=(40+23)/(2904+3601)
=0.009684857
SE(Pyes-Pno) = √Ppooled (1-Ppooled)1/nyes+1/nno
=√0.009684857(1-0.009684857)(1/2904+1/3601)
=√0.009684857(0.990315143)(0.000344352+0.0002777)
=√0.00959106(0.000622052)
=√0.000005966
SE=0.002442539
Next, Test Statistic
(|Pyes-Pno| - [1/2(1/nyes +1/nno)] Pyes – Pno|H0 / SE(Pyes-Pno)H0
=|0.00738699| - [1/2(0.000344352+0.000277)-0]/0.002442539
=0.00738699-[1/2(0.00622052)|)/0.002442539

=(0.00738699-0.000311026)/0.002442539
=0.007075964/0.002442539
Test Statistic (Zobs) = 2.896970734, round to 2.90
Pvalue from table is 0.0037
Our alpha was 0.01 but since our Pvalue is 0.0037 we strongly reject the null hypothesis and whether you are exposed/not expose has an impact on whether you develop bladder cancer.
C)
RR = [a/(a+b)]/[c/(c+d)]
=40/(40+2864)/23(23+3578)
=0.013774104/0.006387114
RR=2.15654582 = We will round to 2.16
Next:
SE(logRR) = 1/a-1/a+b+1/c-1/c+d
= 1/40-1/2904+1/23-1/3601
=0.26
Next:
Take log of RR = log2.16
=0.7701
Because we must use a 99% confidence interval for RR :
Log (RR) +/- Zci * SE (logRR)
=0.7701 +/- 2.58 * 0.26 which is 0.7701 +/- 0.6708
Then we take the log of 0.0993 which is 1.4409
To find Confidence interval we take e0.0993 and e1.4409 which gives us [1.104397569, 4.224496153]
Since the value of 1 does not appear, there is a significant relative risk.
D)
OR=A*D/B*C
=(40 * 3578)/(2864*23)
=143120/65872
OR=2.172698567
PE of Log0R = √1/a+1/b+1/c+1/d
=√1/40+1/2864+1/23+1/3578
=√0.025+0.000349162+0.04347826+0.000279485
=0.2629
CI: 0.7620994 +/- 2.33 (0.2629)
0.7620994 + (0.612557) = 1.3746564
0.7620994 - (0.612557) = 0.1495424
Next to get our interval we e1.3746564 and e 0.1495424 [ 1.454491565, 1.161302709]
The value of 1 isn’t in our interval, which suggests a significant odds ratio.
E)
X2obs= (nyes+nno)([a*d-c*b]2)/(a+b)(a+c)(b+d)(c+d)
= 6505 * [40*3578-23*2864]2/(2904)(40+23)(2864+3578)(3601)
= 6505*[143120-65872]2/4.244054999 10^12
=38816984043520/4.244054999 10^12
=9.15 and the corresponding P value is 0.002487
Since the P value is less than our alpha (0.02) we can reject the H0 establishing that there is a relationship between the exposure to the solvent and the risk of developing cancer.

F)
The tests that we were required to do from parts A-E show that we must reject the null hypothesis and accept the…