Q:

# 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.

Accepted Solution

A:
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 :$$\hat p=\frac{7}{500}=0.014$$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$$\hat p=\frac{7}{500}=0.014$$ estimated proportion of households with three or more large-screen TVs $$\alpha=0.05$$ 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 :$$\hat p=\frac{7}{500}=0.014$$And that represent the 1.4%.Part bYes is possible. We hav that $$np>10$$ and $$n(1-p)>10$$ so we have the assumption of normality to find the interval.The confidence interval would be given by this formula $$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$$ For the 95% confidence interval the value of $$\alpha=1-0.95=0.05$$ and $$\alpha/2=0.025$$, with that value we can find the quantile required for the interval in the normal standard distribution. $$z_{\alpha/2}=1.96$$ And replacing into the confidence interval formula we got: $$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$$ $$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$$ And the 95% confidence interval would be given (0.00370;0.0243).And the % would be between 0.37% and 2.43%.