Question

A researcher claims that 19% of US adults are smokers. In a random sample of 250...

A researcher claims that 19% of US adults are smokers. In a random sample of 250 US adults 55 were found to be smokers. Use a 0.05 significance level to test the researcher’s claim.

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Homework Answers

Answer #1

To test :

H0 : The percentage of US adult smoker is 0.19 , i.e , p = 0.19

H1 :  The percentage of US adult smoker is 0.19 , i.e , p 0.19

Here , n = 250 and alpha = 0.05

The sample proportion , = 55 / 250

= 0.22

The test statistic , z = ( - p) / {p(1-p)/n}

= 0.03 / 0.0248

= 1.21

We know , at alpha = 0.05 , z(critical) = 1.96

Therefore, z < z(critical) and we know that , we fail to reject the null hypothesis when z is less than the critical value.

Therefore , we fail to reject the null hypothesis here and conclude that 19% of US adults are smokers.

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