Question

When trying to hire managers and executives, companies sometimes verify the academic credentials described by the applicants. One company that performs these checks summarized their findings for a six month period. Of the 81 applicants whose credentials were checked, 13 lied about having a degree. A) Find the proportion of applicants who lied about having a degree and find out the standard error.P hat:SE:B) consider these data to be a random sample of credentials from a large collection of similar applicants. Give a 90% confidence interval for the true proportion of applicants who lie about having a degree.Lower bound:Upper bound:

Answer 1

Solution;

`\begin{gathered} n=81 \\ x\text{ \lparen mean\rparen=13} \end{gathered}`

To determine Proportion;

`p=(x)/(n)`
`\begin{gathered} p=(13)/(81) \\ p=0.1604938 \end{gathered}`

To determine the standard error;

`\begin{gathered} SE=\frac{standard\text{ deviation}}{√(n)}=(√(npq))/(√(n)) \\ where\text{ q=1-p} \end{gathered}`
`SE=(3.30357)/(9)=0.36706`

To determine 90% confidence interval;

`\begin{gathered} C.I=x\pm z(standard\text{ error}) \\ where\text{ z is the z score at 90}\%\text{ confidence interval.} \end{gathered}`
`\begin{gathered} C.I=13\pm1.645(0.36706) \\ C.I=13\pm0.6038137 \\ 13-0.6038136<\text{ 13+0.6038136} \\ 12.3962<13.6038\text{ to 4 d.p} \end{gathered}`

Lower bound is 12.3962

Upper bound is 13.6038