.A national survey of companies included a question that asked whether the company
had at least one bilingual telephone operator. The sample results of 90 companies follow (Y
denotes that the company does have at least one bilingual operator; N denotes that it does
not).

Use this information to estimate with 95% confidence the proportion of the population that does have at least one bilingual operator.

Answer:

The 95% confidence the proportion of the population that does have at least one bilingual operator is (0.2359, 0.4307).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

90 companies. 30 have at least one billingual operator.

This means that [tex]n = 90, \pi = \frac{30}{90} = 0.3333[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3333 - 1.96\sqrt{\frac{0.3333*0.6667}{90}} = 0.2359[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3333 + 1.96\sqrt{\frac{0.3333*0.6667}{90}} = 0.4307[/tex]

The 95% confidence the proportion of the population that does have at least one bilingual operator is (0.2359, 0.4307).

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