Question

Project Number 9 Statistical analysis of radiation data recorded by an ionization chamber

Answer 1

3.Statistical analysis:

Data Set 1(nC)

Mean	Median
457.09091	472.0

Data Set 2(nC)

Mean	Median
400.27273	407.0

b) difference between mean and median for data set 1= 14.90909

difference between mean and median for data set 2 = 6.72727

One-Sample T: data set(1)

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
33	457.1	58.9	10.3	(436.2, 478.0)

μ: mean of data set(1)

Test

Null hypothesis	H₀: μ = 472
Alternative hypothesis	H₁: μ ≠ 472

T-Value	P-Value
-1.45	0.156

We have an evidence that P-Value is greater than 0.05 to accept the null hypothesis . 95% confidence interval for mean is (436.2, 478.0).

One-Sample T: data set(2)

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
33	400.3	58.5	10.2	(379.5, 421.0)

μ: mean of data set(2)

Test

Null hypothesis	H₀: μ = 407
Alternative hypothesis	H₁: μ ≠ 407

T-Value	P-Value
-0.66	0.514

We have an evidence that P-Value is greater than 0.05 to accept the null hypothesis . 95% confidence interval for mean is (379.5, 421.0).

c)

	data set 1	data set2
Mean	457.09091	400.27273
Median	472	407
Max	537	498
Min	305	264
Range	232	234
Variance	3469.7727	3426.267
Standard deviation	58.904777	58.534324

d)

These two samples would likely be independent because the ionization chamber was placed at two different distances from gamma radiation source. We Know something about the distribution of values in the first sample doesn't inform you about the distribution of values in the second.So, the sample measurements for both the data sets are dependent of the field size because if distance change , sample measurements will change.

f)

correlation coefficient of data set 1 is 0.936. There are strongly positive correlation between field size and sample measurments of data set1.

correlation coefficient of data set 2 is 0.953.There are strongly positive correlation between field size and sample measurments of data set 2.

g)

data set 2 vs field size data set(2) = 302.2 + 5.768 field size 500 s R-Sq R-Sq(adj) 18.0565 90.8% 90.5% 450 400 data set(2) 350 300 250 0 5 10 25 30 35 15 20 field size