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Morgarella [4.7K]
3 years ago
9

The Corporate Lawyer, a magazine for corporate lawyers, reported that out of 200 firms with employee stock ownership plans, 150

indicated that the primary reason for setting up the plan was tax related. Develop a 90% confidence interval for the population proportion of all such firms with this as the primary motivation. (2 points) (Answers should be in percentage form. Round answers to 2 decimal places)
Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

The 90% confidence interval for the population proportion of all such firms with this as the primary motivation is (69.96%, 80.04%).

Step-by-step explanation:

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

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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

The Corporate Lawyer, a magazine for corporate lawyers, reported that out of 200 firms with employee stock ownership plans, 150 indicated that the primary reason for setting up the plan was tax related.

This means that n = 200, \pi = \frac{150}{200} = 0.75

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.}{2} = 0.975, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.645\sqrt{\frac{0.75*0.25}{200}} = 0.6996

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.645\sqrt{\frac{0.75*0.25}{200}} = 0.8004

As percent:

0.6996*100% = 69.96%

0.8004*100% = 80.04%.

The 90% confidence interval for the population proportion of all such firms with this as the primary motivation is (69.96%, 80.04%).

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