Question

You weigh a random sample of adult golden retrievers and get the following results: 55, 64, 58, 61, 69, 64, 59, 69, 72, and 65. Which answer gives a 98% confidence interval for the mean of the population

Answer 1

Answer:

The 98% confidence interval for the mean of the population is (59, 68.2).

Step-by-step explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation.

Sample mean:

$\overline{x} = \frac{55+64+58+61+69+64+59+69+72+65}{10} = 63.6$

Sample standard deviation:

$s = \sqrt{\frac{(55-63.6)^2+(64-63.6)^2+(58-63.6)^2+(61-63.6)^2+(69-63.6)^2+(64-63.6)^2+(59-63.6)^2+(69-63.6)^2+(72-63.6)^2+(65-63.6)^2}{10}} = 5.142$

Confidence interval:

We have the standard deviation for the sample, and thus, we use the t-distribution.

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 = 10 - 1 = 9

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$. So we have T = 2.821

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.821\frac{5.142}{\sqrt{10}} = 4.6$

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 63.6 - 4.6 = 59

The upper end of the interval is the sample mean added to M. So it is 63.6 + 4.6 = 68.2.

The 98% confidence interval for the mean of the population is (59, 68.2).