Question

Two populations are considered: cell phone users and nonusers. n1 = 98 cell phone users and n2=102 non-users are randomly selected and checked for brain cancer. Is the proportion of brain cancer different in these 2 populations? Perform the Z test for comparison of proportions (Normal Theory method) to test

	Cell-users	Non-users	Total
Brain cancer	18	7
No brain cancer	80	95
Total	98	102

H_o: p_{bc cell-user} =p_{bc non-cell-user}i.e. H_o: p₁=p₂

H_a: p_{bc cell-user} ≠ p_{bc non-cell-user} H_a: p₁ ≠ p₂

Find the proportion of non-cell-users who have developed brain cancer.

	a.	.0686
	b.	The correct answer does not appear as one of the choices.
	c.	.875
	d.	.035
	e.	7

Answer 1

Find the proportion of non-cell-users who have developed brain cancer.

the proportion of non-cell-users who have developed brain cancer = 7/(7+18) = 7/25 = 0.28

So correct choice is "The correct answer does not appear as one of the choices"

Let's use minitab:

Step 1: Click on Stat >>> Basic Statistics >>>2 Proportions...

Step 2: Select Summarized data

Fill the given information

Look the following picture ...

Then click on Option:

Look the following image:

2 Proportions - Options Confidence level: 95.0 Test difference: 0.0 Alternative: notequa 反IJsepooled.estimateofp.fortesti Help OK Cancel

Then click on OK again click on Ok

So we get the following output

From the above output

z = 0.2.46 , p-value = 0.014

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.014 < 0.05 so we used first rule.

That is we reject null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to say that the sample data indicates the proportion of brain cancer different in these 2 populations