Question

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Here are the data:

Computer software was used to analyze the data. The output included the value ${\mathrm{X}}^2=0.435$

Which of the following may we conclude, based on the chi-square test results?

a. There is convincing evidence of an association between wiring configuration and the chance that a child will develop some form of cancer.

b. HCC either causes cancer directly or is a major contributing factor to the development of cancer in children.

c. Leukemia is the most common type of cancer among children.

d. There is not convincing evidence of an association between wiring configuration and the type of cancer that caused the deaths of children.

e. There is convincing evidence that HCC does not cause cancer in children.

Short answer

Option(d) is correct since there is adequate evidence to support the assertion of an association.

Step-by-step solution

Given information

We need to conclude the results based on chi-square test.

	Leukemia	Lymphoma	Other	Total
HCC	$52$	$10$	$17$	$79$
LCC	$84$	$21$	$31$	$136$
Total	$136$	$31$	$48$	$215$

Simplify

The test-statistic value is shown beneath the two-way table:

$X^2=0.435$

The P-value is the chance of getting the test statistic's value, or a number that is more extreme.
The P-value is the number (or interval) in Table C's column title that contains the row's t-value

$df=(r-1)(c-1)=(2-1)\times (3-1)=2$

$P>0.25$

The null hypothesis is rejected if the P-value is less than or equal to the significance level:
$P>0.05\Rightarrow \mathrm{}\mathrm{Fail}\mathrm{to}\mathrm{reject}{\mathrm{H}}_0$

Answer (d) is correct since there is adequate evidence to support the assertion of an association.