Question

The UN National Survey on Drug Use and Health reports that 4.8% of persons 12 and older regularly use prescription pain relievers for non-medical purposes. Suppose a random sample of 900 persons who are 12 and older taken this year found 54 using them for non-medical purposes.

1. Construct a confidence interval around the sample proportion at 99% Confidence Level.

B. Test the hypothesis that the proportion of non-medical pain reliever use among 12 year and older has increased. Use the critical value approach and 1% level of significance.\

Answer 1

(a) p = 54/900 = 0.06

n = 900

z = 2.576

The 99% confidence interval will be:

= p ± z*(√p(1-p)/n

= 0.06 ± 2.576*(√0.06(1-0.06)/900

= (0.0396, 0.0804)

The 99% confidence interval is between 0.0396 and 0.0804.

(b) The hypothesis being tested is:

H0: p = 0.048

Ha: p > 0.048

The test statistic, z = (p̂ - p)/√p(1-p)/n

z = (0.06 - 0.048)/√0.048(1-0.048)/900

z = 1.68

The p-value is 0.0461.

The critical value is -2.33

Since 1.68 < 2.33, we cannot reject the null hypothesis.

Therefore, we cannot conclude that the proportion of non-medical pain reliever use among 12 years and older has increased.

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