Use the traditional method in testing the hypothesis in the problems below:

a. State the hypotheses and identify the claim.

b. Find the critical value(s)

c. Find the test value

d. Make the decision

e. Summarize the result

A nurse was hired by a governmental ecology agency to investigate the impact of a lead smelter on the level of lead in the blood of children living near the smelter. Ten children were chosen at random from those living near the smelter. A comparison group of seven children was randomly selected from those living in an area relatively free from possible lead pollution. Blood samples were taken from the children and lead levels determined. The following are the results (scores are in micrograms of lead per 100 milliliters of blood):

Children living near smelter: 18 16 21 14 17 19 22 24 15 18

Children living in unpolluted area: 9 13 8 15 17 12 11 – – –

Analyze the result using two-tailed test with ∝= 0.10. 

Since we have small sample sizes and they were taken from normally distrubution population(most likely it is), Using T-test is appropriate

Null hypothesis: “u{\scriptscriptstyle 1} = u{\scriptscriptstyle 2}”

Null hypothesis: “u{\scriptscriptstyle 1} \not = u{\scriptscriptstyle 2}”

Where “u{\scriptscriptstyle 1}, u{\scriptscriptstyle 2}” – near the smelter mean and far from smelter sample means respectively

T statistic is building the following way:

“T={\frac {u{\scriptscriptstyle 1}-u{\scriptscriptstyle 2}} {\sqrt{{\frac {s^{2}{\scriptscriptstyle 1}} n}+ {\frac {s^{2}{\scriptscriptstyle 2}} m}}}}” , where “s^{2}{\scriptscriptstyle 1},s^{2}{\scriptscriptstyle 2}” – near the smelter variance and far from smelter sample means respectively. n, m – near the smelter sample size and far from smelter sample size respectively

“u{\scriptscriptstyle 1}={\frac {18+16+…+18} {10}}= 18.4”

“s^{2}{\scriptscriptstyle 1} ={\frac {(18-18.4)^{2}+…+(18-18.4)^{2}} 9}= {\frac {90.4} {9}}=10”

“u{\scriptscriptstyle 2}={\frac {9+…+11} {7}}= 12.4”

“s^{2}{\scriptscriptstyle 1} =6*{\frac {(13-12.4)^{2}+…+(11-12.4)^{2}} 6}={\frac {61.2} {6}} = 10.2”

Now we can calculate value of the t-statistic:

“T={\frac {18.4-12.4} {\sqrt{1+1.45}}} = 3.83”

critical value is “T{\scriptscriptstyle {cr}}(0.05;15)”, 15 – degree freedom(n+m-2), 0.05 – half of the “\alpha”

If T є (“-T{\scriptscriptstyle {cr}}, T{\scriptscriptstyle {cr}})” then we fail to reject null hypothesis

“T{\scriptscriptstyle {cr}}(0.05;15)=1.753”

Or T-value is greater than critical, so we reject the null hypothesis.

There are statistically significant evidence that living near the smelter does impact the blood of children