Question

A dog breeder would like to know how many Dalmatian puppies are typically born in a litter. He conducts some research and selects a random sample of 101 Dalmatian birth records. He examines each birth record and identifies the number of puppies that were born in the litter. The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter and a standard deviation of 2.1 puppies per litter. He would like to estimate, with 99% confidence, the mean number of puppies born per litter for all Dalmatian births. Which of the following is the correct interval?
a. 6.2 ± 2.576(0.21)
b. 6.2 ± 2.576(2.1)
c. 6.2 ± 2.626(0.21)
d. 6.2 ± 2.626(2.1)
e. The confidence interval

Answer 1

Answer:

c. 6.2 ± 2.626(0.21)

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 = 101 - 1 = 100

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.626

The confidence interval is:

$\overline{x} \pm M$

In which $\overline{x}$ is the sample mean while M is the margin of error.

The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter.

This means that $\overline{x} = 6.2$

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.626\frac{2.1}{\sqrt{101}} = 2.626(0.21)$

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

Thus, the confidence interval is:

$\overline{x} \pm M = 6.2 \pm 2.626(0.21)$

And the correct answer is given by option c.