Question

a sample of 20 managers was taken, and they were asked whether or not they usually take work home. the responses of these managers are given below, where \yes\ indicates they usually take work home and
o\ means they do not.
no yes yes yes no yes no
no no no no yes yes no
yes yes no no no yes yes
construct a 95% confidence interval for the percentage p of all managers who take work home.
round your answers to two decimal places.
lower bound:
upper bound:

Explanation:

Step1: Count the number of "Yes" responses

There are 10 "Yes" responses out of \(n = 20\) managers. So, the sample proportion \(\hat{p}=\frac{10}{20}=0.5\)

Step2: Find the critical value

For a 95% confidence interval, the significance level \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). The critical value \(z_{\alpha/2}\) is \(z_{0.025}\). From the standard normal table, \(z_{0.025} = 1.96\)

Step3: Calculate the margin of error

The formula for the margin of error \(E\) for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
Substitute \(\hat{p}=0.5\), \(n = 20\), and \(z_{\alpha/2}=1.96\)

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Step4: Calculate the confidence interval

The confidence interval for the proportion \(p\) is \(\hat{p}-ESubstitute \(\hat{p}=0.5\) and \(E = 0.220\)
Lower bound: \(0.5-0.220 = 0.280\) (as a proportion) or \(28.0\%\)
Upper bound: \(0.5 + 0.220=0.720\) (as a proportion) or \(72.0\%\)

Answer:

Lower bound: \(28.0\%\)
Upper bound: \(72.0\%\)