Distribution of dog heights

[superwolf]

Homework Statement

The distribution of heights of a certain breed of terrier dogs has a mean height of 72 cm and a standard deviation of 10 cm, whereas the distribution of heights of a certain breed of poodles has a mean height of 28 cm with a stanndard deviation of 5 cm. Assuming that the sample means can be measured to any degree of accuracy, find the probability that the sample mean for a random sample of heights of 64 terriers exceeds the sample mean for a random sample of heights of 100 poodles by at most 44.2 cm.

Homework Equations

Central Limit Theorem:

$$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$$

The Attempt at a Solution

Can I simply calculate the chance that the average sample height of the terriers terriers does not exceed (28 cm + 44.2 cm) = 72.2 cm ?

 

[Mark44]

I'm pretty sure you need to be using the two-sample t test. Your null hypothesis is
H₀: $\mu_T \leq \mu_P$
Your alternate hypothesis is
H_a: $\mu_T > \mu_P$
You'll be looking at the probability in the right-hand tail of the distribution.

 

[statdad]

No, a t-test won't give a probability that one sample mean is larger than another - it will indicate whether one population mean is larger than another, and we already know that is the case.

Use of the CLT for both is correct: think these steps.

1) use the CLT to find the distribution for the mean height of the specified group of terriers - call the sample mean Tbar (I don't have time to use Latex now)

2) use the clt to find the distribution for the mean height of the poodles - call the sample
mean Pbar

3) Since these are different breeds of dogs, you know the distributions of the sample breeds are INDEPENDENT, so you know how to find the distribution of Tbar - Pbar

The probability you need to calculate, based on Step 3, is

P(Tbar - Pbar <=44.2) (average height of terrier mean exceeds average height of poodles by at most 44.2)