Question

In a survey of 630 randomly selected U.S. companies, 535 reportedthat they performed drug testing on their employees and/or applicantslast year. At the 1% significance level, is there sufficient evidence toconclude that the percentage of U.S. firms that drug-tested last yearexceeds the previous year's figure of 74%? Formulate and carry outthe one-population z-test.Mean = 300Your answer should include:a) a statement of the hypothesesb) the critical value for the testc) the value of the test statisticd) a statement of conclusionTo get you started, here is the null hypothesis:H.p = 0.74

Answer 1

Step-by-step explanation:

Given the following:

`\begin{gathered} \alpha=0.05 \\ p=0.74 \\ \mu=300 \end{gathered}`

a)

`\begin{gathered} H_0\colon P=0.1 \\ H_1\colon P<0.1 \end{gathered}`

`\begin{gathered} z=\frac{\hat{P}-P}{\sqrt[]{(P(1-P))/(\mu)}} \\ \\ =\frac{0.09-0.1}{\sqrt[]{(0.1\mleft(0.9\mright))/(300)}}=-0.58 \end{gathered}`

b) The critical value is -1.645

Also

`\begin{gathered} p<-0.58 \\ =0.28 \end{gathered}`

c) Conclusion:

We fail to reject the test, since there is no enough evidence to conclude that the proportion of wrong tests is less than 10%