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Sampling and Data 9 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. PROBLEM 5 This problem is worth 1 point An approximate 90% confidence interval for the average number of cars among households in the city goes from _____________________ to _______________________.

OpenStudy (ybarrap):

\[\mu \pm z\alpha\] you already have average and standard deviation, what is "z" corresponding to 90th confidence interval?

OpenStudy (ybarrap):

The z that gives 90% confidence interval is 1.645: http://www.wolframalpha.com/input/?i=inverse+normal+z%3D1.645

OpenStudy (ybarrap):

\[1.8\pm1.645*1.3\] \[Lower Range: 1.8-1.645*1.3\] \[Upper Range: 1.8+1.645*1.3\] Does this make sense?

OpenStudy (anonymous):

@ybarrap What about this? An approximate 99% confidence interval for the percent of city households that have no car goes from _____________________% to ______________________%.

OpenStudy (anonymous):

6,14% to 13,86% ?

OpenStudy (anonymous):

Confidence Interval For Proportions

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