## Related questions with answers

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the 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.The drug OxyContin (oxycodone) is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 227 subjects were treated with OxyContin and 52 of them developed nausea (based on data from Purdue Pharma L.P.). Use a 0.05 significance level to test the claim that more than 20% of OxyContin users develop nausea. Does the rate of nausea appear to be too high?

Solution

VerifiedGiven:

$\alpha=0.05$

$x=52$

$n=227$

Given claim: More than 20% or 0.20.

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

$H_0:p=0.20$

$H_a:p>0.20$

The sample proportion is the number of successes divided by the sample size:

$\hat{p}=\dfrac{x}{n}=\dfrac{52}{227}\approx 0.2291$

Determine the value of the test-statistic:

$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.2291-0.20}{\sqrt{\dfrac{0.20(1-0.20)}{227}}}\approx 1.10$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, when the null hypothesis is true. Determine the P-value using the normal probability table in the appendix.

$P=P(Z>1.10)=1-P(Z<1.10)=1-0.8643=0.1357$

If the P-value is smaller than the significance level $\alpha$, then reject the null hypothesis:

$P>0.05\Rightarrow \text{ Fail to reject } H_0$

There is not sufficient evidence to support the claim that more than 20% of OxyContin users develop nausea and thus the rate of nausea appears to be too high.

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