I am not asking for you to solve number 3 ONLY number 4. There are two things I\

Question

I am not asking for you to solve number 3 ONLY number 4. There are two things I'm looking for.

3. 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 large screen 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. 4. (This continues exercise 3.) Among the sample households, 121 had no car, 172 had one car, and 207 had two or more cars. Estimate the percentage of households in the town with one or more cars; attach a standard error to the estimate. If this is not possible, explain why not.

Explanation / Answer

Solution:

3)

a) proportion = p = 7/500 = 0.014 = 1.4%
b)=0.05, two tail distribution z = 1.96

95% confidence interval for all 25000 household with 3+ large screen tv

=p ± z * (p(1-p)/n)

=0.014±1.96 * (0.014(1-0.014)/500)

= (0.0037, 0.0243) or (.37%, 2.43%)

4)

75.8%, 1.92%

(172+207)/500 = 379/500 = 75.8%
(sqrt 500) * 0.4238 / 500 = 0.0192

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