Mathematics: A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were born, and 342 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?

3 over 2Step-by-step explanation:Given the expressionUsing the law of indices to simplifyHence option A is correct 3 over 2...

A clinical trial tests a method designed to increase

Ответ:

jeremytheprimerapper

15.08.2021

Mathematics

The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.

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 $1 - \frac{\alpha}{2}$ .

In the study 380 babies were born, and 342 of them were girls.

This means that $n = 380, \pi = \frac{342}{380} = 0.9$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so Z = 2.575 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.8604$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.9396$

As percentages:

0.8604*100% = 86.04%.

0.9396*100% = 93.96%.

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Ответ:

Meap12345678910

02.01.2020

Mathematics

3 over 2

Step-by-step explanation:

Given the expression

$\frac{3^{-4} \times 2^3 \times 3^2}{2^4 \times 3^{-3}}$

Using the law of indices to simplify

$= \frac{3^{-4} \times 3^2 \times 2^3}{2^4 \times 3^{-3}}\\= \frac{3^{-4+2} \times 2^3}{2^4 \times 3^{-3}}\\= \frac{3^{-2} \times 2^3}{2^4 \times 3^{-3}}\\= \frac{3^{-2}}{3^{-3}} \times \frac{2^3}{2^4} \\= 3^{-2+3} \times 2^{3-4}\\= 3^1 \times 2^{-1}\\= 3 \times \frac{1}{2}\\= \frac{3}{2}$

Hence option A is correct 3 over 2

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