Question

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 1

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=(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%.

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=(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 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 :

`\hat p=(7)/(500)=0.014`

And that represent the 1.4%.

Part b

Yes 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{(\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{(0.014(1-0.014))/(500)}=0.00370`

`0.014 + 1.96 \sqrt{(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%.