Question

Can you show me how to figure out A using the formula BY HAND? (not with excel)

The research division of a large chemical company tried each of three termite repellants on 200 wooden stakes driven into ran- dom locations. Two years later, the number of infected stakes 17-8 was as follows: Ci- Treatment Т, T2 T1 Infected? 48 18 26 Yes No 29 174 152 182 co Note that this experiment was designed to have exactly the same number of stakes (200) for each treatment. Thus the relative fre- quency for each treatment is 200/600 estimate any underlying population proportion. So the sampling differs from the simple random sample of Table 17-3. Nevertheless, the standard x2 test still remains valid. When used this way, it is often called a x2 test with fixed marginal totals, or a x2 test of homogeneity (Are the various treatments homogeneous, i.e., similar in terms of infection rate?) 1/3, and does not really a. Calculate x2 and the p-value. b. Are the treatments discernibly different at the 5% level?

Answer 1

Answer:

• $\chi^2$ test of Homogeneity:

The given contingency table is as below:

Infected ?	Treatment			Row Total
	T1	T2	T3
Yes	26	48	18	92
No	174	152	182	508
Column Total	200	200	200	600

For answering the questions (a) and (b), we have to perform the six steps of hypotheses testing given as below:

Step 1: Set the null and alternative hypotheses

The null and alternative hypotheses can be stated as below:

H₀: The three treatments are homogeneous.

H₁: The three treatments are not homogeneous.

Step 2: Determine the appropriate statistical test

The test statistic for $\chi^2$ test of homogeneity is

$\chi^2=\sum_{all\ cells}\frac{(f_o-f_e)^2}{f_e}$

with degrees of freedom = (number of rows - 1) x (number of columns - 1)

where, f_o = Observed frequency

f_e = Expected frequency =

RT X CT N

RT = row total, CT = column total and N = Grand total.

Step 3: Set the decision rule

For the given value of , rules for acceptance or rejection of null hypothesis are as below:

= 0,05

If reject the null hypothesis, otherwise, do not reject the null hypothesis.

Xcal > Xcritical

By using the standard Chi-square table, the critical $\chi^2$ value is

đủ 05 2 = 5.9915

where degrees of freedom = (number of rows - 1) x (number of columns - 1) = (2-1) x (3-1) = 2.

Step 4: Computing the test statistic

Expected frequency for cell (1, 1) (from the above given table) can be calculated as below:

fe11 = RT X CT y = frac92 x 200600 = 30.67

Similarly, the expected frequencies for other cells can be calculated.

The contingency table with observed and expected frequencies is as below:

Infected ?	Treatment			Row Total
	T1	T2	T3
Yes	26 (30.67)	48 (30.67)	18 (30.67)	92
No	174 (169.33)	152 (169.33)	182 (169.33)	508
Column Total	200	200	200	600

Computation for chi-square statistic:

fo (Observed fre)	fe (Expected fre)	(f. -fe)
26	30.67	0.711082
48	30.67	9.792269
18	30.67	5.234069
174	169.33	0.128795
152	169.33	1.773631
182	169.33	0.948024
Total		I 40 - b)! = 18.5879

So,

ved = [ (40–10? 18.5879

Step 6: Arrive at statistical conclusion

At 95% confidence level (
= 0,05
), the critical value obtained from the chi-square table is
đủ 05 2 = 5.9915
. And the calculated value is
Xcal = 18.5879
, which is greater than the critical value and falls in rejection region. Hence the null hypothesis is rejected and alternative hypothesis is accepted.

There is not enough evidence to indicate that the treatments are homogeneous. So, we can conclude that the treatments are different at 5% level of significance.

Ans (a):

Form the above calculations,

Chi-square statistic is given as:

Xcal = 18.5879

And the corresponding p-value = 0.0000919 which is same as Zero.

Ans (b):

For the Step 6 in above given answer, we can say that, the treatments are different at 5% level of significance.