A study suggests that about 80% of undergraduate students are working at a job. Researchers plan to do a new study and use the new data to make a z interval on of students who are working at a job. They plan on using a confidence level of 90% and they want the margin of error to be no more than 1%. Let p represent the proportion of students who are working at a job and assume p 0.80, what is the smallest sample size required to obtain the desired margin of error?

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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

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

In this problem, we have that:

[tex]\pi = 0.8[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

What is the smallest sample size required to obtain the desired margin of error?

This is n when [tex]M = 0.01[/tex]

So

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

[tex]0.01 = 1.645\sqrt{\frac{0.8*0.2}{n}}[/tex]

[tex]0.01\sqrt{n} = 1.645\sqrt{0.8*0.2}[/tex]

[tex]\sqrt{n} = \frac{1.645\sqrt{0.8*0.2}}{0.01}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.8*0.2}}{0.01})^{2}[/tex]

[tex]n = 4329.6[/tex]

Rounding up

The smallest sample size required to obtain the desired margin of error is 4330

angelimpat23 angelimpat23 · Answer 2 · 2022-02-24T20:39:03+01:00

angelimpat23 angelimpat23

24-02-2022

Answer:

4330

Step-by-step explanation:

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