1. Statistics Sampling Distributions
Chapter 6
Example Problems

2. Sample Proportions

3. Sample Proportions
Answer: To find the sample proportions, look at each pair of values. So we have
3,3
The sample proportion would be the number of odd values (since this is what the problem asked) / total number of values = 2 odd values / 2 total values = 1

4. Sample Proportions We can continue this with And so on….until we get
3, 4 would be 1 odd / 2 total = 1/2 or 0.5
3,11 would be 2 odd / 2 total = 1
And so on….until we get
0 odds: 4,4 1 odd: 3,4; 4,3; 4,11; 11,4 2 odd: 3,3; 3,11; 11,3; 11,11

5. Sample Proportions
Now to find the probability remember it is the number of possible odd / total. So if you look at the previous 0 there are 1 out of 9 total.
Next, to find the probability that we would get one odd and one even would be the second row and there are 4 groups out of 9 total or 4/9
And finally to find even,even we actually could use the complement 1 – (odd,odd and odd,even) = 1 – 1/9 – 4/9 = 9/9 – 1/9 – 4/9 = 4/9

6. Normal Distribution Example
Question: Assume that women's heights are normally distributed (this is the key to use the normal z table) with a mean given by mean = 63.4 inches and a standard deviation = 1.8 inches.
a.  If a woman is randomly selected, find the probability that her height is between inches and 63.9 inches.

7. Normal Distribution Example
Answer: Find the z value with the formula z = (x – mean) / standard deviation for the lower value of x = 62.9 and upper value x = 63.9
Low z = (62.9 – 63.4) / 1.8 = (rounded to two decimals)
Upper z = (63.9 – 63.4) / 1.8 = 0.28 (rounded to two decimal places)

8. Normal Distribution Example
So “between” would be the area between z = and z = 0.28

9. Normal Distribution Example
The area from the bottom of the normal curve up to z = is found in the Negative z scores table. Look up -0.2 in the row and .08 in the column to get
The area from the bottom of the normal curve up to z = 0.28 is found in the Positive z scores table. Look up 0.2 in the row and .08 in the column to get
Since this is the area between, if you subtract these two values you get the area between
= .2206

10. Normal Distribution Example 2
Question: Cans of a certain beverage are labeled to indicate that they contain 20 oz. The amounts in a sample of cans are measured and the sample statistics are n = 42 and sample mean = oz. If the beverage cans are filled so that population mean = oz (as labeled) and the population standard deviation is = oz (based on the sample results), find the probability that a sample of 42 cans will have a mean of oz or greater. Do these results suggest that the beverage cans are filled with an amount greater than oz?

11. Normal Distribution Example 2
Answer: Since the sample size is greater than 20 we can approximate with a normal distribution using the formula

12. Normal Distribution Example 2
We look up the z = 2.38 in the Normal Table to get the area = However, we are looking at the “amount greater than oz” and thus will be 1 – = which gives are probability that the sample of 42 cans will have a mean of oz or greater.
We look to compare with 0.05 (standard if not given) and see that this probability being way less certainly makes this particular event rate if the true mean given is oz.