Mathematics: A survey of 1295 student loan borrowers found that 460 had loans totaling more than $20,000 for their undergraduate education. Give a 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education. (Give answers accurate to 3 decimal places.)

A survey of 1295 student loan borrowers found - id12955993820, jaimefashion12

Ответ:

holytreesaroy

23.01.2022

The 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education is (0.324, 0.386)

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1295, \pi = \frac{460}{1295} = 0.355$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so Z = 2.327 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.355 - 2.327\sqrt{\frac{0.355*0.645}{1295}} = 0.324$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.355 + 2.327\sqrt{\frac{0.355*0.645}{1295}} = 0.386$

The 98% confidence interval for the proportion of all student loan borrowers who have loans of $20,000 or more for their undergraduate education is (0.324, 0.386)

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Ответ:

narwhalebearp5871i

12.03.2023

Mathematics

Total price $11.15

Step-by-step explanation:

Tip Total:$ 1.35

Total (Bill + Tip):$ 10.32

Sales tax: $0.83

Cost/Price before ST: $10.32

Total Cost/Price including ST: $11.15

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