Question

Dogsled drivers, known as mushers, use several different breeds of dogs to pull their sleds. One proponent of Siberian Huskies believes that sleds pulled by Siberian Huskies are faster than sleds pulled by other breeds. He times 47 teams of Siberian Huskies on a particular short course, and they have a mean time of 5.2 minutes. The mean time on the same course for 39 teams of other breeds of sled dogs is 5.5 minutes. Assume that the times on this course have a population standard deviation of 1.4 minutes for teams of Siberian Huskies and 1.1 minutes for teams of other breeds of sled dogs. Let Population 1 be sleds pulled by Siberian Huskies and let Population 2 be sleds pulled by other breeds. Step 1 of 2 : Construct a 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs

Answer 1

Answer:

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.  

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Siberian Huskies:

Sample of 47, mean of 5.2 minutes, standard deviation of 1.4. So

$\mu_1 = 5.2$

$s_1 = \frac{1.4}{\sqrt{47}} = 0.2042$

Others:

Sample of 39, mean of 5.5 minutes, standard deviation of 1.1. So

$\mu_2 = 5.5$

$s_2 = \frac{1.1}{\sqrt{39}} = 0.1761$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 5.2 - 5.5 = -0.3$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.2042^2+0.1761^2} = 0.2692$

Confidence interval:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = zs$

In which s is the standard error. So

$M = 1.96(0.2692) = 0.5276$

The lower end of the interval is the sample mean subtracted by M. So it is -0.3 - 0.5276 = -0.8276.

The upper end of the interval is the sample mean added to M. So it is -0.3 + 0.5276 = 0.2276

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).