A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The 90% confidence interval around this sample mean is

The 90% confidence interval around this sample mean is between $127.51 and $203.29.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the students 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 = 17 - 1 = 16

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of ). So we have T = 1.7459

The margin of error is:

M = T*s = 1.7459*21.70 = 37.89

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 165.40 - 37.89 = $127.51

The upper end of the interval is the sample mean added to M. So it is 165.40 + 37.89 = $203.29

The 90% confidence interval around this sample mean is between $127.51 and $203.29.

answer: there are x+184 pages in olivia's book.

step-by-step explanation:

let the total number of pages in the book which olive has already read be x.

then if olive has 184 pages left to read, then the total number of pages are in olive's book= pages she has read+pages she has left to read

therefore, the total number of pages in olivia's book=

hence, there are x+184 pages in olivia's book.