ESSENT STATISTICS&MYSTATISTICSLAB&SPSS MNL Chapter 8.3, Problem 15BSC

Question:

Testing Claims About Proportions. In Exercise, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution (as described in Part 1 of this section).

Cell Phones and Cancer In a study of 420,095 Danish cell phone users, 135 subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today). Test the claim of a somewhat common belief that such cancers are affected by cell phone use. That is, test the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of 0.0340% for people who do not use cell phones. Because this issue has such great importance, use a 0.005 significance level. Based on these results, should cell phone users be concerned about cancer of the brain or nervous system?

Solution:

In the study, among 420,095 randomly selected Danish cell phone users, 135 subjects developed cancer of the brain or nervous system. So we have, n = 420,095, x = 135

The claim is that the cell phone users develop cancer of brain or nervous system at a rate that is different from the rate of 0.0340% of people who do not use cell phones. Since the null hypothesis is the hypothesis of equality, our claim becomes the alternative hypothesis and the equality becomes the null hypothesis.

Thus the hypothesis to be tested is, H0: p=0.000340 VS H1: p ≠ 0.000340

As per the statement of the problem, the significance level is 0.005. The estimate of sample proportion is, Using these values, obtain the value for test statistics as,

z= \frac{\widehat{p}-p}{\sqrt{\frac{pq}{n}}}
= \frac{0.000321-0.000340}{\sqrt{\frac{(0.000340)(1-0.00034)}{420095}}}
=\frac{-0.000019}{0.0000284}
=-0.6679

To obtain the critical region, use the Table A-2. The snapshot of the table is as below.

Table A-2 Standard Normal (z) Distribution : Cumulative Area From The LEFT
z.00.01.02.03.04.05.06.07.08.09
-2.9.0019.0018.0018.0017.0016.0016.0015.0015.0014.0014
-2.8.0026.0025.0024.0023.0023.0022.0021.0021.0020.0019

The value of level of significance is 0.005, so for two tailed test, the level of significance is 0.0025. Look for the probability value 0.0025 in the table A-2. The value 0.0025 occurs in table A-2 along the row -2.8 and column 0.01. Thus the required value is -2.81. But since this is a two-tailed test, the critical value is ± 2.81

To obtain the p-value, use the same table A-2 for negative z-scores since test statistic is negative.

Table A-2 Standard Normal(z) Distribution : Cumulative Area From The LEFT
z.00.01.02.03.04.05.06.07.08.09
-0.6.2743.2709.2676.2643.2611.2578.25546.2514.2483.2451
-0.5.3085.3050.3015.2981.2946.2912.2877.2843.2810.2776

Obtain the probability value for the z-score -0.66. That means look for the row -0.6 and -0.66 column 0.06. The value obtained is 0.2546. Since this is a two-tailed test, the p-value is the double the probability value obtained. So the p-value is 2×0.2546 = 0.5092

Since the p-value > 0.005 and also the test statistic z=-0.66 is greater than the left critical value -2.81. Thus the test statistic lies within the non-critical region. So we accept the null hypothesis Ho and so in turn reject our claim that the cell phone users develop cancer of brain or nervous system at a rate that is different from the rate of 0.0340% of people who do not use cell phones. The study thus reviles that the cell phone users need not be concerned about cancer of the brain and nervous system.

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