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2 years ago

PLEASE ANSWER WILL GIVE BRAINLIEST IF CORRECT!!!

Mathematics

2 answers:

swat322 years ago

8 0

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Airida [17]2 years ago

5 0

Answer:

x = 118 units

Step-by-step explanation:

tan 50° = x/99

x = 99(tan 50°)

x = 117.9

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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, a

Bas_tet [7]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

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 interval $1-\alpha$ , 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:

In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.

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