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

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

Explanation:

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

Sample mean:

`\overline{x} = (55+64+58+61+69+64+59+69+72+65)/(10) = 63.6`

Sample standard deviation:

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 - (1 - 0.98)/(2) = 0.99`. So we have T = 2.821

The margin of error is:

`M = T(s)/(√(n)) = 2.821(5.142)/(√(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).