Question

In: Statistics and Probability

A university dean is interested in determining the proportion of students who receive some sort of...

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 225 students and finds that 45 of them are receiving financial aid. Using a 95% confidence interval, what is the upper limit of the confidence interval to estimate the true proportion of students who receive financial aid.

Using a 95% confidence interval, what is the upper limit of the confidence interval to estimate the true proportion of students who receive financial aid.

Expert Solution

Solution :

Given that,

n = 225

x = 45

Point estimate = sample proportion = = x / n = 45/225=0.2

1 - = 1- 0.2 =0.8

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z / 2 * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ( $\sqrt$ ((0.2*0.8) /225 )

= 0.0523

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2-0.0523 < p <0.2+ 0.0523

0.1477< p < 0.2823

The 95% confidence interval for the population proportion p is : lower limit =0.1477,upper limit=0.2823

orchestra answered 2 years ago

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