# How construct a 95% confidence interval estimate for u and write the confidence statement?College students throw an average of 640 pounds of trash each year. To estimate the amount of trash...

How construct a 95% confidence interval estimate for u and write the confidence statement?

College students throw an average of 640 pounds of trash each year. To estimate the amount of trash discarded by the students at the State University, 36 students were randomly selected. The mean of the sample was 559 pounds with a standard deviation of 159 pounds. Construct a 95% confidence interval estimate for u(average amount of trash that college students throw out each year) and write the confidence statement.

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We are given a sample with `n=36,bar(x)=559,s=159` . We are asked to find a confidence interval with 95% confidence `(alpha=.05)` for the population mean `mu` .

The population mean will reside in an interval created by the sample mean plus/minus a correction factor that takes into account the confidence level and the standard error. The formula to use is:

`bar(x)-t_(alpha/2)(s/sqrt(n))<=mu<=bar(x)+t_(alpha/2)(s/sqrt(n))` Here `s/sqrt(n)` is the standard error while `t_(alpha/2)` takes into account the confidence level.

From a Student's t-distribution table we have d.f.=35 and `alpha/2=.025` . However, for values of `n>29` we can use the z-distribution table. So `t_(alpha/2)=1.96`

Plugging in the given values we get:

`559-1.96(159/sqrt(36))<=mu<=559+1.96(159/sqrt(36))`

`507.06<=mu<=610.94`

**My calculator gives the t-interval as `505.2<=mu<=612.8` . This is due to rounding error when using the tables**

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**We claim with a 95% confidence level that State University students throw away on average between 507 and 611 pounds of trash per year.**

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