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A NHANES report gives data for 654 women aged 20–29 years. The mean BMI of these 654 women was x¯=26.8 . We treated these data as an SRS from a normally distributed population with standard deviation ????=7.5 . (a) Suppose that we had an SRS of just 100 young women. What would be the margin of error for 95% confidence?

Respuesta :

Answer: Margin of error would be 1.47 for 95% confidence.

Step-by-step explanation:

Since we have given that

Mean = 26.8

Standard deviation = 7.5

n = 100

We need to find the margin of error for 95% confidence.

So, z = 1.96

So, the margin of error would be

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{7.5}{\sqrt{100}}\\\\=\dfrac{14.7}{10}\\\\=1.47[/tex]

Hence, margin of error would be 1.47 for 95% confidence.

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