A random sample of 856 births in a state included 426 boys. Construct a $95 %$ confidence interval estimate of the proportion of boys in al births it is believed that among at tirths, the proportion of boys is 0.509 . Do these sample results provide strong evidence against that belier?

Construct a $95 %$ confidence interval estimate of the proportion of boys in al births.

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Answer

The 95% confidence interval estimate of the proportion of boys in all births is $(0.4676, 0.5300)$

Steps

Step 1 ：Calculate the sample proportion of boys: $Sample Proportion = \frac{426}{856} = 0.4988$

Step 2 ：Calculate the standard error: $Standard Error = \sqrt{\frac{0.4988 \cdot (1 - 0.4988)}{856}} = 0.0159$

Step 3 ：Calculate the margin of error: $Margin of Error = 1.96 \cdot 0.0159 = 0.0312$

Step 4 ：Construct the confidence interval: $Confidence Interval = 0.4988 \pm 0.0312 = (0.4676, 0.5300)$

Step 5 ：The 95% confidence interval estimate of the proportion of boys in all births is $(0.4676, 0.5300)$