Question

The National Vital Statistics Reports for November 2011 states that U.S. preterm birth rate for 2010 was about 12%. Preterm birth means premature birth. Suppose that this year a random sample of 500 births has 52 that are preterm. Prompt Using the estimate from the NVS Report for 2011 and the result from this year’s random sample, estimate the percent of preterm births this year with 95% confidence interval. (Be sure to check that a normal model is appropriate and show your work for calculating the 95% confidence in a fashion similar to the examples.)

Answer 1

Answer:

$0.104 - 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.0772$

$0.104 + 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.1308$

The 95% confidence interval would be given by (0.0772;0.1308)

Step-by-step explanation:

For this case we can calculate the estimated proportion like this:

$\hat p =\frac{52}{500}= 0.104$

We can assume that the normal model can be used since we satisfy two conditions:

$np =500*0.104= 52>10$

$n(1-p) = 500*(1-0.104) = 448>10$

The confidence level for this case is 95% of confidence, and the significance is given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the values we got:

$0.104 - 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.0772$

$0.104 + 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.1308$

The 95% confidence interval would be given by (0.0772;0.1308)

Answer 2

Answer:

Therefore, the 95% confidence interval would be given by= (0.0771 , 0.1309)

Step-by-step explanation:

Given : Sample size=n=500

Let, X be the number of preterm.

Here , X=52

The sample proportion is , p=X/n=52/500=0.1040

q= 1 - p = 0.8960

Therefore , the 95% confidence interval estimate for the percent of preterm of births this year is ,

$P\pm Z_{\alpha /2}\sqrt{\frac{pq}{n} }$

$0.1040 \pm Z_{0.05/2}\sqrt{\frac{0.1040 \times 0.8960}{500} } \\\\0.1040*1.96*0.0137$

From normal probability integral table;

$Z_{0.05/2} =1.96\\\\0.1040 \pm 0.0269$

Therefore, the 95% confidence interval would be given by= (0.0771 , 0.1309)