 # Tutorial: Sampling Distribution of the Mean: N = 5

Your researcher friend has considered using sample sizes of only five people. She will ask you to explain the advantages and disadvantages of this plan.  To begin, select the following options: Normal, N=5, Show sample data (no other ‘show’ options).

Click on Draw a sample, and record the value shown for Last mean in the first space below. Then click on Draw a sample nine more times and record each mean below.

How many of your 10 sample means fell outside of the range 470 to 530?

Click on Show obtained means to see all ten means displayed. Notice how tightly clustered the obtained means are compared to the individual scores, and to the distributions of means when N=100.

Click Draw 100 samples to see the means for 100 different samples, each of size N=5.

Click Show sampling distribution of the mean to see how closely the observed sample means match the actual distribution of possible means of size N=5.

In the box below describe how this sampling distribution of the mean (for N=5) compares to the sampling distribution of the mean for N=100.

Q4. What is the probability that a randomly selected sample of N=5 American adults has a mean Life Satisfaction score within 30 points of the population mean?

You can estimate the answer by examining your ten sample means and the displays of 100 sample means with N=5 for each mean. What is your estimate based on these observations?

You may use a table of probabilities for the standardized normal distribution or use the pz converter to convert from a z-score to probability.

#### Show A Hint

We know that the possible means are normally distributed with a mean of 500. If we can find the standard deviation of this distribution, we can find the z score corresponding to 530, and then use the z table or p-z converter to find the probability of observing a sample mean between 500 and 530, and between 500 and 470.

#### Show Another Hint

According to the CLT, the standard error of the mean is equal to the population standard deviation, 100, divided by the square root of the sample size, 5. Thus, the standard error = 100/ (square root of 5) = 100/2.236 = 44.721. Using this standard error of 44.721, you can now convert 470 and 530 (scores that deviate from the mean by 30) into z scores and find their corresponding probabilities.

A detailed solution is in the answer section, but try it on your own before consulting the answer.

4,162 total views,  1 views today