Respuesta :
Answer:
a) [tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic
b) [tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic
c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Part a
Description in words of the parameter p
[tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic
[tex]\hat p[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic
n=1714 is the sample size required
[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Part b
Numerical estimate for p
In order to estimate a proportion we use this formula:
[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.
[tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic
Part c
The confidence interval for a proportion is given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545[/tex]
[tex]0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591[/tex]
And the 95% confidence interval would be given (0.545;0.591).
We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.
