Question

Most people complain that they gain weight during the December holidays. To find out how much, we sample the weights of 23 adults in mid-November and again in early to mid-January. The mean weight change for the sample was a gain of 1.64 lbs., with a standard deviation of the differences of 8.73 lbs. Find a 88% confidence level for the average weight gain.

Answer 1

Answer:

The 88% confidence level for the average weight gain if between -1.30 lbs and 4.58 lbs.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 23 - 1 = 22

88% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 22 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.88}{2} = 0.94$. So we have T = 1.6176

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.6176\frac{8.73}{\sqrt{23}} = 2.94$

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 1.64 - 2.94 = -1.30 lbs

The upper end of the interval is the sample mean added to M. So it is 1.64 + 2.94 = 4.58 lbs

The 88% confidence level for the average weight gain if between -1.30 lbs and 4.58 lbs.