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 :
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%.