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10.106 Public Health and Nutrition Over the last decade, many Americans have been able to stop smoking. However, a recent survey suggests that asthmatic children are more likely to be exposed to second-hand smoke than children without asthma.19 In a random sample of 300 children without asthma, 132 were regularly exposed to second-hand smoke, and in a random sample of 325 children with asthma, 177 were regularly exposed to second-hand smoke. Is there any evidence to suggest that the proportion of children with asthma who are exposed to second-hand smoke is greater than the proportion for children without asthma? Use a 5 0.01.

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Answer:

There is no significant evidence at 0.01 significance level, to suggest that the proportion of children with asthma who are exposed to second-hand smoke is greater than the proportion for children without asthma

Step-by-step explanation:

[tex]H_{0}:[/tex] the proportion of children with asthma who are exposed to second-hand smoke is equal the proportion for children without asthma

[tex]H_{0}:[/tex] the proportion of children with asthma who are exposed to second-hand smoke is greater than the proportion for children without asthma

Test statistic can be calculated using the formula

z=[tex]\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

  • p1 is the proportion of children without asthma who are exposed to second-hand smoke ([tex]\frac{132}{300}=0.440[/tex] )
  • p2 is the proportion of children with asthma who are exposed to second-hand smoke ([tex]\frac{177}{325}=0.545[/tex] )
  • p is the pool proportion of p1 and p2 ([tex]\frac{132+177}{300+325}=0.494[/tex])
  • n1 is the sample size of children without asthma (300)
  • n2 is the sample size of children with asthma (325)

Then z=[tex]\frac{0.545-0.440}{\sqrt{{0.494*0.506*(\frac{1}{300} +\frac{1}{325}) }}}[/tex] ≈1.679

P-value of test statistic is  ≈0.047. Since 0.047>0.01 (significance level) we fail to reject the null hypothesis.

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