# Solved (Free): Researchers at the University of South Florida conducted a study of drug usage of U.S. physicians. The anonymous survey of 5,430 randomly selected physicians

#### ByDr. Raju Chaudhari

Mar 28, 2021

Researchers at the University of South Florida conducted a study of drug usage of U.S. physicians. The anonymous survey of 5,430 randomly selected physicians revealed that 429 experienced substance abuse or drug dependency in their lifetime. Test the hypothesis that more than 5% of US. physicians have used or depended on drugs in their lifetime at a 0.05.

State the null and alternative hypothesis.

Give the 95% Confidence Interval.

Give a conclusion for the confidence interval test.

#### Solution

Given that number of randomly selected physicians $n = 5430$, number of physicians who experienced substance abuse or drug dependency in their lifetime is $X= 429$.

The sample proportion is

$$\hat{p}=\frac{X}{n}=\frac{429}{5430}=0.079$$.

The hypothesis testing problem is
$H_0 : p = 0.05$ against $H_1 : p > 0.05$ ($\text{right-tailed}$)

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned}
which follows $N(0,1)$ distribution.

The significance level is $\alpha = 0.05$.

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.64}$.

The rejection region (i.e. critical region) for the hypothesis testing problem is $\text{Z > 1.64}$.

The test statistic is

 \begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.079-0.05}{\sqrt{\frac{0.05* (1-0.05)}{5430}}}\\ & =9.807 \end{aligned}

The test statistic is $Z =9.807$ which falls $inside$ the critical region, we $\text{reject}$ the null hypothesis.

OR

$p$-value approach:

This is a $\text{right-tailed}$ test, so the p-value is the area to the right of the test statistic ($Z=9.807$). Thus the $p$-value = $P(Z>9.807) =0$.

The p-value is $0$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

Conclusion We conclude that more than 5% of US. physicians have used or depended on drugs in their lifetime at a 0.05 level of significance.

$95$% confidence interval estimate for population proportion

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

$100(1-\alpha)$% confidence interval for population proportion is
 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

The margin of error for proportions is
 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.079*(1-0.079)}{5430}}\\ & =0.007. \end{aligned}

$95$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.079 - 0.007 & \leq p \leq 0.079 + 0.007\\ 0.0718 & \leq p \leq 0.0862. \end{aligned}

Thus, $95$% confidence interval estimate for population proportion is $(0.0862,0.0718)$.

Conclusion

We are 95% confident that the true proportion of US physicians used or depended on drugs in their lifetime lies between $(0.0862,0.0718)$.