Respuesta :
Answer:
[tex]0.0553 - 1.96\sqrt{\frac{0.0553(1-0.0553)}{380}}=0.0323[/tex]
[tex]0.0553 + 1.96\sqrt{\frac{0.0553(1-0.0553)}{380}}=0.0783[/tex]
Step-by-step explanation:
The info given is:
[tex] X= 21[/tex] number of students who said that they planned to work in a rural community
[tex] n= 380[/tex] represent the sample size selected
[tex]\hat p =\frac{21}{380}= 0.0553[/tex] the estimated proportion of students who said that they planned to work in a rural community
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be 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]
Replpacing we got:
[tex]0.0553 - 1.96\sqrt{\frac{0.0553(1-0.0553)}{380}}=0.0323[/tex]
[tex]0.0553 + 1.96\sqrt{\frac{0.0553(1-0.0553)}{380}}=0.0783[/tex]
A mean for estimation is the minimum-maximum variation estimate's C.I. The % of pupils planning to work in a rural community alters between 0.0323 and 0.0783.
Confidence interval:
Let's [tex]p^{}[/tex] represent the sampling fraction of the people who promised to work in a rural area.
Sample size:
[tex]n = 380[/tex]
x: the large number the pupils expected to work in a rural setting
[tex]p^{} = \frac{x}{n} \\\\p^{} = \frac{21}{ 380} = 0.0553\\\\(1- \alpha)\ \ 100\%[/tex]confidence for true proportion:
[tex]( p^{}\ \pm Z_{\frac{\alpha}{2}} \times \sqrt{p^{} \times \frac{(1-p^{})}{n}} ) \\\\[/tex]
For [tex]95\%[/tex]confidence interval:
[tex]\to 1 - \alpha = 0.95[/tex]
When:
[tex]\to \alpha = 0.05[/tex]
Calculating the value of Z by using the table:
[tex]\to Z_{0.025} = 1.96[/tex]
When the [tex]95\%[/tex] of the confidence interval:
[tex]\to (0.0553 \pm Z_{0.025} \times \sqrt{(0.0553 \times \frac{(1- 0.0553)}{380}})\\\\\to (0.0553 - Z_{0.025} \times \sqrt{(0.0553 \times \frac{(1- 0.0553)}{380})},0.0553 + Z_{0.025} \times \sqrt{(0.0553 \times \frac{(1- 0.0553)}{380}))}\\\\[/tex]
by solving the value we get:
[tex]\to ( 0.0323 , 0.0783 )[/tex]
- We are [tex]95\%[/tex] sure that the true proportion of students planning to work in a rural community is between [tex]0.0323[/tex] and [tex]0.0783[/tex].
- That is we are [tex]95\%[/tex] sure that the percentage of students planning to work in a rural community is between [tex]3.23\%[/tex] and [tex]7.83\%[/tex].
Find out more about the Confidence interval here:
brainly.com/question/2396419
