Answer to Question #250502 in Statistics and Probability for smilynne

Question

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

2. 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.

Answer 1

Let us first summarize our data,

$n_1=10$

$n_2=7$

The sample means are given by the formula,

$\bar{x}_i=(\sum (x))/n_i$

For sample 1,

$\bar{x}_1=184/10=18.4$

and for sample 2

$\bar{x}_2=85/7=12.1428571$

The sample variances are given by the formula,

$S^2_i=(\sum(x^2)-(\sum(x))^2/n_i)/(n_i-1)$

For sample 1

$S_1^2=(3476-(184)^2/10)/9=10.0444444$

sample 2

$S_2^2=(1093-(85)^2/7)/6=10.1428567$

The hypotheses tested in this scenario is,

$H_0:\mu_1=\mu_2$

$Against$

$H_1:\mu_1\not=\mu_2$

A two tailed test implies that the claim is that the two population means are the same. To perform this test, we use the students' t distribution as below.

The test statistic is given as,

$t^*=(\bar{x_1}-\bar{x}_2)/\sqrt{Sp^2(1/n_1+1/n_2)}$ where $Sp^2$ is the pooled sample variance given by the formula,

$Sp^2=((n_1-1)S_1^2+(n_2-1)S^2_2)/(n_1+n_2-2)$

$Sp^2=((9*10.0444444)+(6*10.1428567))/15=10.0838093$

So,

$t^*=(18.4-12.142871)/\sqrt{10.083(1/10+1/7)}=6.2571429/\sqrt{2.44892511}$

Hence,

$t^*=3.998(3\space decimal\space places)$

$t^*$ is compared with the t distribution critical value at $\alpha=0.1$ with $(n_1+n_2-2)=(10+7-2)=15$

Critical value $t_{\alpha/2,15}=t_{0.1/2,15}=t_{0.05,15}=1.753$

The null hypothesis is rejected if $t^*\gt t_{0.05,15}$

Since $t^*=3.998\gt t_{0.05,15}=1.753$ we reject the null hypothesis.

We conclude that sufficient evidence exist to indicate a significant difference between level of lead in the blood of children living near the smelter and those living in an unpolluted area.

Children living near the smelter have significantly higher levels of lead in the blood than those living in the unpolluted area.