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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 390 babies were​ born, and 312 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?

Respuesta :

Answer:Yes

Step-by-step explanation:

Given

n=390 x=312

[tex]\hat{p}=\frac{312}{390}=0.8[/tex]

Confidence level=99 %

[tex] Z_{\frac{\alpha }{2}}=2.575[/tex]

Standard error(S.E.)=[tex]\sqrt{\frac{\hat{p}\left ( 1-\hat{p}\right )}{n}}[/tex]

S.E.=[tex]\sqrt{\frac{0.8\times 0.2}{390}}[/tex]

S.E.=0.0202

Confidence interval

[tex]p\pm \left [ z_{\frac{\alpha }{2}}\cdot S.E.\right ][/tex]

[tex]0.8 \pm 0.0521[/tex]

[tex]\left ( 0.7479,0.8521 \right )[/tex]

Since 0.5 does not lie in interval therefore method appear to be effective

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