Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in the town with three or more largescreen TVs is estimated as ; this estimate is likely to be off by or so. (b) If possible, find a 95%-confidence interval for the percentage of all 25,000 households with three or more large-screen TVs. If this is not possible, explain why not.

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 TVsPart aThe 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 bYes 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%.