How do you solve D. ?

Question

Dogs weigh more than cats, on average. Suppose a random sample of 40 dogs gives a sample mean of lbs with a standard deviation of  and a random sample of 50 cats gives a sample mean of with a standard deviation of .

 

a.​What is the mean of the sampling distribution of ?

b.​What is the standard deviation of the sampling distribution of ? Verify the 10% condition?

c.​What is the shape of the sampling distribution of ? Show how you know.

d.​What is the probability that a sample of 40 dogs weighs 50 lbs more than a sample of 50 cats? Show your work.

Answer 1

I think you need to use the facts you get from the from answering a., b. and c.

For example:

a. The mean of any distribution is the average of all your data points.

b part1. The standard deviation of any distribution is the average amount of variation in your data;

or on average, how far each data point is from the mean. It can NOT be negative & if it =0, all your data points are the same.

b part2. The 10% condition says that the sample size should be less than or equal to 10% of the population size in order to safely make the assumption that a set of Bernoulli trials is independent.

c. Since the sample sizes are 40 & 50, we can use the 'central limit theory', which says "as n increases, the sampling distribution of the mean is approximately normal. An n≥30 is often given as a rule of thumb for a sufficiently large sample.

So, to answer d: Use what you know about testing a null hypothesis for a variable that follows a normal distribution. I find the wording unclear, but I think it's asking you to find the probability that:

(mean dog wt)+50>(mean cat wt). I suggest you let x=dog, and y=cat. Substituting variable symbols may help you see how to solve the problem. Hope you find this helpful.