Respuesta :
Answer:
a) The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :
[tex]\hat p=\frac{7}{500}=0.014[/tex]
And that represent the 1.4%.
b) And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.
Step-by-step explanation:
Data given and notation
n=500 represent the random sample taken
X=7 represent the households with three or more large-screen TVs
[tex]\hat p=\frac{7}{500}=0.014[/tex] estimated proportion of households with three or more large-screen TVs
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)
z would represent the statistic (variable of interest)
p= population proportion of households with three or more large-screen TVs
Part a
The percentage of households in the town with three or more largescreen TVs is estimated as :
The best estimation for the population proportion is :
[tex]\hat p=\frac{7}{500}=0.014[/tex]
And that represent the 1.4%.
Part b
Yes is possible. We hav that [tex]np>10[/tex] and [tex]n(1-p)>10[/tex] so we have the assumption of normality to find the interval.
The confidence interval would be 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]0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370[/tex]
[tex]0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243[/tex]
And the 95% confidence interval would be given (0.00370;0.0243).
And the % would be between 0.37% and 2.43%.
