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.

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 :

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

estimated proportion of households with three or more large-screen TVs

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 :

And that represent the 1.4%.

Part b

Yes is possible. We hav that and so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

And replacing into the confidence interval formula we got:

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

c

Step-by-step explanation:

I just know