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serious [3.7K]
2 years ago
15

Someone HELP PLEASEEE

Mathematics
1 answer:
olganol [36]2 years ago
3 0

Answer:

300

Step-by-step explanation:

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3/10=0.3

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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335335 babies were​ born
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Answer:

The 99​% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

Step-by-step explanation:

Confidence Interval for the proportion:

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 = 335, \pi = \frac{268}{335} = 0.8

99% confidence level

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)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.7437

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.8563

For the percentage:

Multiplying the proportions by 100.

The 99​% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

7 0
3 years ago
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