Respuesta :
Solution :
The sample proportion [tex]$(\hat p)=\frac{24}{1000}$[/tex]
= 0.024
90% confidence interval
The standard deviation = [tex]$\sqrt{\frac{\hat p(1-\hat p)}{n}$[/tex]
= [tex]$\sqrt{\frac{0.024(1-0.024)}{1000}$[/tex]
= 0.00484
z = 1.645 for 90% CI
Upper band = 0.024 + (0.00484 x 1.645 ) = 0.03196
Lower band = 0.024 - (0.00484 x 1.645 ) = 0.01603
Therefore, the 90% CI is (0.016, 0.032)
99% confidence interval
z = 2.576 for 99% CI
Upper band = 0.024 + (2.576 x 0.00484 ) = 0.0365
Lower band = 0.024 - (2.576 x 0.00484 ) = -0.0115
Therefore, the 99% CI is (0.0115, 0.0365)
Using the z-distribution, it is found that:
- The 90% confidence interval to estimate the proportion of people who have type 2 diabetes is (0.016, 0.032).
- The 99% confidence interval to estimate the proportion of people who have type 2 diabetes is (0.012, 0.036).
- The relationship is that a higher confidence level leads to a wider interval.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
24 out of 1000 people who were surveyed had type 2 diabetes, hence:
[tex]n = 1000, \pi = \frac{24}{1000} = 0.024[/tex]
90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so [tex]z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.024 - 1.645\sqrt{\frac{0.024(0.976)}{1000}} = 0.016[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.024 + 1.645\sqrt{\frac{0.024(0.976)}{1000}} = 0.032[/tex]
The 90% confidence interval to estimate the proportion of people who have type 2 diabetes is (0.016, 0.032).
99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.024 - 2.575\sqrt{\frac{0.024(0.976)}{1000}} = 0.012[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.024 + 2.575\sqrt{\frac{0.024(0.976)}{1000}} = 0.036[/tex]
The 99% confidence interval to estimate the proportion of people who have type 2 diabetes is (0.012, 0.036).
The relationship is that a higher confidence level leads to a higher value of z, making the interval wider.
A similar problem is given at https://brainly.com/question/15850972
