Question

Dogs vary in size a great deal more than cats do. Suppose the mean weight of dogs in a certain city is 28 pounds with a standard deviation of 14 pounds, and the mean weights of cats is 9.5 pounds with a standard deviation of 2 pounds. Cat weights are roughly Normally distributed, but dog weights are skewed right. Suppose you take a random sample of 10 dogs and 8 cats. Which of the following is an accurate description of the center, spread, and shape of the sampling distribution of the difference in the mean weight dogs and the mean weight of cats?

a. Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape approximately Normal
b. Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape non-Normal
c. Mean = 18.5 pounds, standard deviation - 14.1 pounds, shape approximately Normal
d. Mean = 18.5 pounds, standard deviation = 14.1 pounds, shape non-Normal
e. The difference of means is 18.5 pounds and the shape is non-Normal, but the standard deviation cannot be determined since the sample size is small and the distribution of dog weights is non-Normal.

Answer 1

b. Mean = 18.5 pounds, standard deviation = 4.48 pounds, shape non-Normal

Explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes 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 = (\sigma)/(√(n))`.

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction of samples:

When two samples are subtracted, we have that the mean is the difference between the means and the standard deviation is the square root of the sum of the variances.

The mean weight of dogs in a certain city is 28 pounds with a standard deviation of 14 pounds. Sample of 10

This means that
`\mu_(D) = 28, s_D = (14)/(√(10)) = 4.4272`

Mean weights of cats is 9.5 pounds with a standard deviation of 2 pounds. Sample of 8.

This means that
`\mu_(C) = 9.5, s_(C) = (2)/(√(8)) = 0.7071`

Which of the following is an accurate description of the center, spread, and shape of the sampling distribution of the difference in the mean weight dogs and the mean weight of cats?

The mean is:

`\mu = \mu_(D) - \mu_(C) = 28 - 9.5 = 18.5`

The standard deviation(spread) is:

`s = \sqrt{s_(D)^2+s_(C)^2} = √(4.4272^2+0.7071^2) = 4.48`

Normal distribution subtracted by non-normal(sample of cats is less than 30, so wont be normal), will be non-normal.

The correct answer is given by option b.