Question

In a large city, the city supervisor wants to find the average number of aluminum cans that each family recycles per month. So, she surveys 18 families and finds that these 18 families recycle an average of 140 cans per month with a standard deviation of 26 cans per month. Find the 90 % confidence interval for the mean number of cans that all of the families in the city recycle per month.

Answer 1

Answer:

The 90% onfidence interval for the mean number of cans that all of the families in the city recycle per month is between 129.34 and 150.66 cans per month

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.74

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.74\frac{26}{\sqrt{18}} = 10.66$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 140 - 10.66 = 129.34 cans per month

The upper end of the interval is the sample mean added to M. So it is 140 + 10.66 = 150.66 cans per month.

The 90% onfidence interval for the mean number of cans that all of the families in the city recycle per month is between 129.34 and 150.66 cans per month