Question

Give an example, in your field of study or other fields of interest, in which we would want to find the distribution of either

1. Sampling Distribution of X

2. Sampling Distribution of the Difference between 2 Means

3. Sampling Distribution of Proportions

(You only need to provide one example of each scenario.)

Answer 1

1)

If X has a distribution with mean μ, and standard deviation σ, and is approximately normally distributed or n is large,

then is approximately normally distributed with mean μ and standard error

Example:-

μ=20, and σ=5. Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P(> 22)?

So the probability that the sample mean will be >22 is the probability that Z is > 1.6 We use the Z table to determine this:

P( > 22) = P(Z > 1.6) = 0.0548.

2)

It often becomes important to compare two population means. Knowledge of the sampling distribution of the difference between two means is useful in studies of this type. It is generally assumed that the two populations are normally distributed.
EXAMPLE:-
In a study of annual family expenditures for general health care, two populations were surveyed with the following results:
Population 1: n₁ =40   x₁_bar =$346
Population 1: n₂ =35   x₂_bar =$300
If the variances of the populations are σ₁² = 2800 and  σ₂²= 3250, what is the probability of obtaining sample results (x₁_bar - x₂_bar) as large as those shown if there is no difference in the means of the two populations?

Solution:-
Write the given information

n₁ =40, x₁_bar =$346, σ₁² = 2800

n₂ =35, x₂_bar =$300, σ₂²= 3250

Find the Z score

  A value of Z = 3.6 gives an area of .9998. This is subtracted from 1 to give the probability

P (z > 3.6) = .0002Complete the answer
The probability that (x₁_bar - x₂_bar)  is as large as given is .0002.

3)

The Sampling Distribution of Proportion measures the proportion of success, i.e. a chance of occurrence of certain events, by dividing the number of successes i.e. chances by the sample size ’n’. Thus, the sample proportion is defined as p = x/n. The sampling distribution of proportion obeys the binomial probability law if the random sample of ‘n’ is obtained with replacement. Such as, if the population is infinite and the probability of occurrence of an event is ‘π’, then the probability of non-occurrence of the event is (1-π). Now consider all the possible sample size ‘n’ drawn from the population and estimate the proportion ‘p’ of success for each. Then the mean (p) and the standard deviation (σp) of the sampling distribution of proportion can be obtained as:

p = mean of proportion
π = population proportion which is defined as π = X/N, where X is the number of elements that possess a certain characteristic and N is the total number of items in the population.
σp = standard error of proportion that measures the success (chance) variations of sample proportions from sample to sample

n= sample size, If the sample size is large (n≥30), then the sampling distribution of proportion is likely to be normally distributed.The following formula is used when population is finite, and the sampling is made without the replacement
EXAMPLE:-

A random sample of 100 students is taken from the population of all part-time students in the United States, for which the overall proportion of females is 0.6.

There is a 95% chance that the sample proportion (p-hat) falls between what two values?

First note that the distribution of p-hat has mean p = 0.6, standard deviation

and a shape that is close to normal, since np = 100(0.6) = 60 and n(1 – p) = 100(0.4) = 40 are both greater than 10.The Standard Deviation Rule applies: the probability is approximately 0.95 that p-hat falls within 2 standard deviations of the mean, that is, between 0.6 – 2(0.05) and 0.6 + 2(0.05). There is roughly a 95% chance that p-hat falls in the interval (0.5, 0.7) for samples of this size.