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Mathematics 27 Online
OpenStudy (anonymous):

A simple random sample of 400 households is taken in a large city, and the number of cars in each sampled household is noted. The average number of cars in the sampled households is 1.8 and the SD is 1.3. Among the sampled households, 10% had no car. An approximate 90% confidence interval for the average number of cars among households in the city goes from _____________________ to _______________________. and An approximate 99% confidence interval for the percent of city households that have no car goes from _____________________% to ______________________%.

OpenStudy (anonymous):

90% is +-1.645 is correct ?

OpenStudy (anonymous):

SE= 1.3 ?

OpenStudy (nurali):

take the sample mean of 1.8, and adjust it by the error: E = Z(standard deviation)/sqrt(400). Z is the zscore of 90%

OpenStudy (nurali):

so: 1.8 +- 1.645(1.3/sqrt(400))

OpenStudy (anonymous):

ok thanks .. and An approximate 99% confidence interval for the percent of city households that have no car goes from _____________________% to ______________________%. ?

OpenStudy (anonymous):

interval from -2.575 to +2.575

OpenStudy (anonymous):

Confidence Interval For Proportions

OpenStudy (anonymous):

http://www.mccallum-layton.co.uk/stats/ConfidenceIntervalCalcProportions.aspx Confidence Interval: ±3.86 Range for the true population proportion 6.14% to 13.86% .. is correct ?

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