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Statistics Sampling Distributions

Chapter 6 Example Problems

Sample Proportions

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

Sample Proportions • We can continue this with 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

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

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 62.9 inches and 63.9 inches.

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 = -0.28 (rounded to two decimals) – Upper z = (63.9 – 63.4) / 1.8 = 0.28 (rounded to two decimal places)

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

Normal Distribution Example • The area from the bottom of the normal curve up to z = -0.28 is found in the Negative z scores table. Look up -0.2 in the row and .08 in the column to get 0.3897. • 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 0.6103. • Since this is the area between, if you subtract these two values you get the area between • 0.6103 - 0.3897 = .2206

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 = 20.04 oz. If the beverage cans are filled so that population mean = 20.00 oz (as labeled) and the population standard deviation is = 0.109 oz (based on the sample results), find the probability that a sample of 42 cans will have a mean of 20.04 oz or greater. Do these results suggest that the beverage cans are filled with an amount greater than 20.00 oz?

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

z

x

n

20 . 04 20 . 00 0 . 109 42

2 . 38

Normal Distribution Example 2 • We look up the z = 2.38 in the Normal Table to get the area = 0.9913. However, we are looking at the “amount greater than 20.00 oz” and thus will be 1 – 0.9913 = 0.0087 which gives are probability that the sample of 42 cans will have a mean of 20.04 oz or greater. • We look to compare with 0.05 (standard if not given) and see that this probability 0.0087 being way less certainly makes this particular event rate if the true mean given is 20.00 oz.

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Chapter 6 Example Problems

Sample Proportions

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

Sample Proportions • We can continue this with 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

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

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 62.9 inches and 63.9 inches.

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 = -0.28 (rounded to two decimals) – Upper z = (63.9 – 63.4) / 1.8 = 0.28 (rounded to two decimal places)

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

Normal Distribution Example • The area from the bottom of the normal curve up to z = -0.28 is found in the Negative z scores table. Look up -0.2 in the row and .08 in the column to get 0.3897. • 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 0.6103. • Since this is the area between, if you subtract these two values you get the area between • 0.6103 - 0.3897 = .2206

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 = 20.04 oz. If the beverage cans are filled so that population mean = 20.00 oz (as labeled) and the population standard deviation is = 0.109 oz (based on the sample results), find the probability that a sample of 42 cans will have a mean of 20.04 oz or greater. Do these results suggest that the beverage cans are filled with an amount greater than 20.00 oz?

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

z

x

n

20 . 04 20 . 00 0 . 109 42

2 . 38

Normal Distribution Example 2 • We look up the z = 2.38 in the Normal Table to get the area = 0.9913. However, we are looking at the “amount greater than 20.00 oz” and thus will be 1 – 0.9913 = 0.0087 which gives are probability that the sample of 42 cans will have a mean of 20.04 oz or greater. • We look to compare with 0.05 (standard if not given) and see that this probability 0.0087 being way less certainly makes this particular event rate if the true mean given is 20.00 oz.

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