Respuesta :
Answer:
[tex]0.104 - 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.0772[/tex]
[tex]0.104 + 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.1308[/tex]
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:
[tex] \hat p =\frac{52}{500}= 0.104[/tex]
We can assume that the normal model can be used since we satisfy two conditions:
[tex] np =500*0.104= 52>10[/tex]
[tex]n(1-p) = 500*(1-0.104) = 448>10[/tex]
The confidence level for this case is 95% of confidence, and the significance is given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Replacing the values we got:
[tex]0.104 - 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.0772[/tex]
[tex]0.104 + 1.96\sqrt{\frac{0.104(1-0.104)}{500}}=0.1308[/tex]
The 95% confidence interval would be given by (0.0772;0.1308)
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 ,
[tex]P\pm Z_{\alpha /2}\sqrt{\frac{pq}{n} }[/tex]
[tex]0.1040 \pm Z_{0.05/2}\sqrt{\frac{0.1040 \times 0.8960}{500} } \\\\0.1040*1.96*0.0137[/tex]
From normal probability integral table;
[tex]Z_{0.05/2} =1.96\\\\0.1040 \pm 0.0269[/tex]
Therefore, the 95% confidence interval would be given by= (0.0771 , 0.1309)
