Oct. 13 - Sampling Distribution of X-Bar(Pt. 1; The Central Limit Theorem)

True or false: The mean of the sampling distribution of x̄ equals μ only when n > 30. - False - True

- False. The mean of the sampling distribution of x̄ equals μ for all sample sizes.

True or false: If a random sample is taken, the shape of the sampling distribution of x̄ is closer to Normal for smaller sample sizes than for larger sample sizes. - True - False

- False. Actually the opposite is true‑the shape of the sampling distribution of x̄ is closer to Normal for larger sample sizes than for smaller sample sizes.

If the shape of the population distribution is non‑Normal, what is the shape of the histogram of data in a sample? - Always Normal regardless of sample size - Approximately Normal for n large and unpredictable for n small - Cannot be predicted for any sample size

- Approximately Normal for n large and unpredictable for n small. If the sample size is large, the shape of the histogram of data in a sample will be approximately the same as the population distribution and gets closer to the population distribution as sample size increases. If sample size is small, we cannot predict the shape.

Computing Probability Example 4: Closing prices of stocks have a right-skewed distribution with μ = 26 and σ = 20. What is the probability that the mean price of a random sample of n = 32 closing stocks is less than $15?

- .0009 1. Draw and label Normal curve. 2. Compute appropriate z-score: z = (x̄ - μ) / (σ / √(n)) = (15 - 26) / (20/√(32)) = -3.11 3. Look up z-score in table to probability: P(x̄ < 15) = P(z < -3.11) = .0009

Computing Probability Example 2: Weights of bottles of chocolate milk are Normally distributed with μ = 1.0875 and σ = 0.015. What is the probability that the mean of a random sample of eight bottles of chocolate milk exceeds 1.1lbs?

- .0091 1. Draw and label Normal curve. 2. Compute appropriate z-score: z = (x̄ - μ) / (σ / √(n)) = (1.1 - 1.0875) / (0.015/√(8)) = 2.36 3. Look up z-score in table to probability: P(x > 1.1) = P(z > 2.36) = 1 - P(z < 2.36) = 1 - .9909 = .0091

Computing Probability Example 1: The BYU Creamery sells bottles of chocolate milk with a mean weight of 1.0875lbs (x̄) and a standard deviation of 0.015lbs (σ). The weights of these bottles are Normally Distributed. What is the probability that a randomly selected bottle weighs more than 1.1lbs?

- .2033 1. Draw and label Normal curve. 2. Compute appropriate z-score: z = (x̄ - μ) / (σ) = (1.1 - 1.0875)/0.015 = 0.83 3. Look up z-score in table to probability: P(x > 1.1) = P(z > 0.83) = 1 - P(z < 0.83) = 1 - .7967 = .2033

If a large population has mean, μ = 40 and standard deviation, σ = 12, what is the standard deviation of the sampling distribution of x̄ created from all possible samples of size n = 9? - cannot be determined since n=9 is too small - 12 - 40 - 9 - 4 - 1.33

- 4 The standard deviation of the sampling distribution of x̄ = σ/√(n) regardless of sample size and shape of the population, so σ/√(n) = 12/√(9) = 4

If a large population has mean, μ = 40 and standard deviation, σ = 12, what is the mean of the sampling distribution of x̄ created from all possible samples of size n = 9? - 1.33 - 12 - 4 - 9 - 40 - cannot be determined since n=9 is too small

- 40. The mean of the sampling distribution of x̄ always exactly equals μ, so the mean equals μ=40.

If the shape of the population distribution is Normal, what is the shape of the sampling distribution of x̄? - Approximately Normal for n large and unpredictable for n small - Always Normal regardless of sample size - Cannot be predicted for any sample size

- Always Normal regardless of sample size If the shape of the population distribution is Normal, the shape of the sampling distribution of x̄ will always be Normal.

If the shape of the population distribution is Normal, what is the shape of the histogram of data in a sample? - Cannot be predicted for any sample size - Always Normal regardless of sample size - Approximately Normal for n large and unpredictable for n small

- Approximately Normal for n large and unpredictable for n small. If the sample size is large, the shape of the histogram of data in a sample will be approximately Normal and gets more Normal as sample size increases. If sample size is small, we cannot predict the shape.

If the shape of the population distribution is non‑Normal, what is the shape of the sampling distribution of x̄? - Cannot be predicted for any sample size - Always Normal regardless of sample size - Approximately Normal for n large and unpredictable for n small

- Approximately Normal for n large and unpredictable for n small. If the shape of the population distribution is non‑Normal, the shape of the sampling distribution of x̄ will be approximately Normal if the sample size is large. If 𝑛n is small and the population distribution is skewed, the shape of the sampling distribution of x̄ will be slightly skewed. There are too many shapes for the population distribution to list all possibilities of shapes of the sampling distribution when n is small.

Why could we use the standard Normal table to find a probability on the mean of a random sample of eight bottles weighing more than 1.1 pounds? - Because the population distribution is Normal. - Actually since the sample size is n=1, we cannot apply the Central Limit Theorem and we should not have used the standard Normal table to find the probability. - Because we wanted to find the probability on a randomly selected bottle.

- Because the population distribution is Normal. Since the population distribution is Normal, the sampling distribution of x̄ is also Normal for all sample sizes. Thus, we could use the standard Normal table to find the probability on the mean weight from a random sample of eight bottles.

Why could we use the standard Normal table to find a probability on a randomly selected bottle weighing more than 1.1 pounds? - Actually since the sample size is n=1, we cannot apply the Central Limit Theorem and we should not have used the standard Normal table to find the probability. - Because the population distribution is Normal. - Because we wanted to find the probability on a randomly selected bottle.

- Because the population distribution is Normal. Since the population distribution is Normal, we could use the standard Normal table to find the probability on the weight of a randomly selected bottle.

Why is the sampling distribution of x̄ approximately Normal? - Because the population distribution is Normal - Because the sample is large and random, allowing us to apply the Central Limit Theorem - Actually, the sampling distribution of x̄ is not approximately Normal

- Because the sample is large and random, allowing us to apply the Central Limit Theorem Since the sample size is 32 and the sample is random, we can apply the Central Limit Theorem and say that the sampling distribution of x̄ is approximately Normal.

Why could we use the z-score z = (x̄ - μ) / (σ / √(n)) and the standard Normal table to find the probability on the mean closing price of a random sample of 32 stocks? - Actually, we should not have used the z‑score and the standard Normal table to find the probability. - Because the sample is random and large. - Because the population is Normally distributed.

- Because the sample is random and large. In this case, we apply the Central Limit Theorem. Even though the population is right‑skewed, non‑Normal, the sample is large and random so the sampling distribution of x̄ is approximately Normal allowing us to use z = (x̄ - μ) / (σ / √(n)) and the standard Normal table to find the probability on the sample mean.

Why don't we need to create the sampling distribution of x̄ before computing a probability on x̄? - Because the shape of the sampling distribution is always Normal allowing us to always use the standard Normal table to find a probability on x̄. - Actually, we must always create the sampling distribution of x̄ in order to compute a probability on x̄. - Because we can use facts about the sampling distribution of x̄ to predict the shape, center, and spread of the sampling distribution of x̄.

- Because we can use facts about the sampling distribution of x̄ to predict the shape, center, and spread of the sampling distribution of x̄. While the shape of a sampling distribution of x̄ is not always Normal, it is when either the population shape is Normal or the sample size is large. Further, the mean of the sampling distribution of x̄ always equals μ and the standard deviation of the sampling distribution of x̄ = σ / √(n) These facts allow us to compute a probability on x̄ without creating the sampling distribution of x̄.

After looking up the z‑score in the standard Normal table, why did we subtract the table probability from 1.0? - Because we wanted the probability that a randomly selected bottle weighs less than 1.1 pounds. - Because we wanted the probability that a randomly selected bottle weighs more than 1.1 pounds.

- Because we wanted the probability that a randomly selected bottle weighs more than 1.1 pounds. The standard Normal table gives area on the left or "less than" probabilities. We needed to find the probability that a randomly selected bottle weighs "more than" 1.1 pounds which is area on the right.

Why is the probability found by looking up the z‑score of -3.11 in the standard Normal table the answer? - Because we wanted the probability that the sample mean is less than $15. - Because we wanted the probability that the sample mean is more than $15.

- Because we wanted the probability that the sample mean is less than $15. The standard Normal table gives area on the left or "less than" probabilities. Since we want a "less than" probability, the number found by looking up z = -3.11 is the correct answer.

If a large population has mean, μ = 40 and standard deviation, σ = 12, what is the shape of the sampling distribution of x̄ created from all possible samples of size n = 9? - Normal - Approximately Normal - Cannot be determined since n=9 is too small for case 2 and we do not know whether the population shape is Normal

- Cannot be determined since n=9 is too small for case 2 and we do not know whether the population shape is Normal For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of x̄ to be approximately Normal, the sample size must be large. Neither of those conditions are met in the description of the problem.

What does the Central Limit Theorem allow us to do? - Compute a probability on a sample mean using a Normal curve if the population has a Normal shape - Compute a probability on x̄ using a Normal curve if the sample is large and random - Compute a probability on sample data using a Normal curve when n is large

- Compute a probability on x̄ using a Normal curve if the sample is large and random. Central Limit Theorem says that the shape of the sampling distribution of x̄ is approximately Normal provided the sample size is large and SRS when sampling from a population having any shape. When the distribution of all the x̄'s is approximately Normal, we can use a Normal curve to compute a probability on x̄.

True or false: The standard deviation of the sampling distribution of x̄ = σ regardless of sample size. - False - True

- False The standard deviation of the sampling distribution of x̄ equals 𝜎/√(n), not σ, regardless of sample size.

True or false: When sampling from a non‑Normal population, the shape of the histogram of the data in the sample gets closer to Normal as the sample size increases. - True - False

- False. When sampling from a non‑Normal population, the shape of the sampling distribution of x̄, not the shape of the histogram of the data in the sample, gets closer to Normal as sample size increases.

Central Limit Theorem (CLT / Case 2 Shape of Sampling Distribution of x̄):

- If you take a large SRS of size n from any population, then the sampling distribution of x̄ is approximately Normal. - Shape gets more Normal as n increases; n > 30 is considered large. - CLT allows us to use the Normal curve to compute approximate probabilities on x̄.

Facts about Sampling Distributions of x̄:

- Means are equal for all n. - Standard deviation of x̄ decreases as n increases. - Shape of sampling distribution of x̄ becomes more Normal as n increases.

Do we have to create the sampling distribution of x̄ every time we want to compute probabilities on x̄? - Yes - No

- No No! We can apply the central limit theorem and use the Normal curve to find probabilities!

Can we use a Normal distribution to find a probability on the closing price of an individual stock? - Yes, because we compute the z-score, z = (x̄ - μ) / (σ / √(n)) since we know the values of μ and σ. - No, because the population distribution is right-skewed, not Normal.

- No, because the population distribution is right-skewed, not Normal. We can only compute a probability on an individual when the population distribution is Normal. The population distribution of closing stock price is right skewed, not Normal so we cannot use a Normal distribution to find a probability on the closing price of an individual stock.

If a large population has a Normal shape with a mean, μ = 40 and standard deviation, σ = 12, what is the shape of the sampling distribution of x̄ created from all possible samples of size n = 9? - Normal - Approximately Normal - Cannot be determined since n = 9 is too small.

- Normal. For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of x̄ to be approximately Normal, the sample size must be large. The population is described to have a Normal shape.

Which graph displays the distribution of a sample of closing stock prices? - The histogram of x̄'s given in red on the right - The histogram of data in a sample given in purple in the middle - The histogram of the population given in blue on the left

- The histogram of data in a sample given in purple in the middle. The histogram given in purple in the middle is a histogram of a sample of closing stock prices.

Which graph displays the distribution of all closing stock prices? - The histogram of the population given in blue on the left - The histogram of x̄'s given in red on the right - The histogram of data in a sample given in purple in the middle

- The histogram of the population given in blue on the left. The population consists of all stocks so the histogram of all closing stock prices is a histogram of the population.

Which graph displays the sampling distribution of x̄? - The histogram of the population given in blue on the left - The histogram of data in a sample given in purple in the middle - The histogram of x̄'s given in red on the right

- The histogram of x̄'s given in red on the right. The sampling distribution of x̄ gives the possible values of x̄ together with how often each occurs. So the histogram of x̄'s displays the sampling distribution of x̄.

According to the Central Limit Theorem, what has an approximate Normal shape if the sample is large and random? - The population distribution - The sampling distribution of x̄ - The histogram of data in sample

- The sampling distribution of x̄. The Central Limit Theorem says that the shape of the sampling distribution of x̄ is approximately Normal if a large random sample is taken. So the Central Limit Theorem has to do with the shape of the sampling distribution of x̄, not the sample and not the population.

What is the definition of the sampling distribution of x̄? - The value of x̄ from a simple random sample. - The values of x̄ from all possible samples of the same size from the same population. - The distribution of values of the response variable in a simple random sample.

- The values of x̄ from all possible samples of the same size from the same population. A distribution of a variable gives the possible values of the variable together with how often each value occurs. So, the sampling distribution of x̄ gives the possible values of x̄ together with how often each value occurs.

True or false: The mean of the sampling distribution of x̄ = μ regardless of sample size. - False - True

- True This is a correct statement.

True or false: The spread of the sampling distribution of x̄ decreases as n increases. - True - False

- True This is a correct statement.

True or false: The standard deviation of the sampling distribution of x̄ = 𝜎/√(n) regardless of sample size and population shape. - False - True

- True. Facts about the mean and the standard deviation of the sampling distribution of x̄ are not part of the Central Limit Theorem and are valid regardless of sample size or shape of the population. The facts do require a simple random sample.

True or false: Whenever the sampling distribution of x̄ is either Normal or approximately Normal and the sample is random, we can use z = (x̄ - μ) / (σ / √(n)) and a Normal distribution to find a probability on a sample mean. - False - True

- True. This is a true statement and will be used repeatedly when we do inference.

True or false: We could compute both a probability on the weight of a randomly selected individual bottle and a probability on the mean weight of a random sample of eight bottles using the standard Normal table because weights of the bottles is Normally distributed. - True - False

- True. Whenever the population distribution is Normal, we can compute a probability on an individual x and a probability on a sample mean using the standard Normal table.

When can we compute a probability on an individual x using a Normal distribution? - When the distribution of data in a sample is Normal - When the population distribution is Normal - When the sampling distribution of x̄ is either Normal or approximately Normal

- When the population distribution is Normal Since we want to compute a probability on an individual x, we can only do this if the population distribution is Normal.

When can we compute a probability on a sample mean, x̄, using a Normal distribution? - When the population distribution is Normal - When the sampling distribution of x̄ is either Normal or approximately Normal - When the distribution of data in a sample is Normal

- When the sampling distribution of x̄ is either Normal or approximately Normal. Since we want to compute a probability on x̄, we need the distribution of x̄'s to be either Normal or approximately Normal.

When is the sample size considered large enough to apply the Central Limit Theorem? - n > 10 - n > 100 - n > 50 - n > 30

- n > 30. n > 30 is considered large enough to apply the Central Limit Theorem although there is no theoretical basis for this number.

What symbol represents the standard deviation of a sample? - σ/√(n) - σ - μ - s - x̄

- s s is the symbol for the standard deviation of a sample.

What symbol represents the mean of a sample? - s - μ - x̄ - σ/√(n) - σ

- x̄ x̄ is the symbol for the mean of a sample.

Which z‑score formula should we use to find a probability on the mean closing price of a random sample of thirty‑two stocks? - Neither - z = (x̄ - μ) / (σ) - z = (x̄ - μ) / (σ / √(n))

- z = (x̄ - μ) / (σ / √(n)) When finding a probability on sample mean, we must use the standard deviation of the sampling distribution of x̄, namely (σ / √(n)), in the denominator. Thus, the appropriate z‑score is z = (x̄ - μ) / (σ / √(n)).

When sampling from a non-Normal population with a large random sample, what z‑score formula should we use to find a probability on? - z = (x̄ - μ) / (σ) - z = (x̄ - μ) / (σ / √(n)) - Neither of the choices are correct.

- z = (x̄ - μ) / (σ / √(n)) When finding a probability on x̄, we must use the standard deviation of the sampling distribution of x̄, namely σ / √(n), in the denominator. Thus, the appropriate z‑score is z = (x̄ - μ) / (σ / √(n)).

Which z‑score formula should we use to find a probability on the mean weight of a random sample of eight bottles? - z = (x̄ - μ) / (σ) - Neither of the choices are correct. - z = (x̄ - μ) / (σ / √(n))

- z = (x̄ - μ) / (σ / √(n)) When finding a probability on x̄, we must use the standard deviation of the sampling distribution of x̄, namely σ / √(n), in the denominator. Thus, the appropriate z‑score is z = (x̄ - μ) / (σ / √(n))

Which z‑score formula should we use to find a probability on an individual bottle? - z = (x̄ - μ) / (σ) - Neither of the choices are correct. - z = (x̄ - μ) / (σ / √(n))

- z = (x̄ - μ) / (σ) When finding a probability on an individual x, we must use the standard deviation of population, namely σ, in the denominator. Thus, the appropriate z‑score is z = (x̄ - μ) / (σ)

What symbol represents the mean of the sampling distribution of x̄? - σ/√(n) - s - μ - σ - x̄

- μ μ is the symbol for the mean of the sampling distribution of x̄.

What symbol represents the mean of a population? - σ - x̄ - s - σ/√(n) - μ

- μ. μ is the symbol for the mean of a population.

What symbol represents the standard deviation of a population? - s - σ/√(n) - σ - μ - x̄

- σ σ is the symbol for the standard deviation of a population.

What symbol represents the standard deviation of the sampling distribution of x̄? - μ - x̄ - σ - σ/√(n) - s

- σ/√(n) σ/√(n) is the symbol for the standard deviation of the sampling distribution of x̄.

Shape of Sampling Distribution of x̄:

Case 1) Population Normal: The shape of the sampling distribution of x̄ is Normal. Case 2) Population Non-Normal: The shape of the sampling distribution of x̄ is approximately Normal when n is large.

The ________ and ___________ _______________ of the Sampling Distribution of x̄ are NOT part of the Central Limit Theorem and are valid for all population distributions and all sample sizes.

Mean Standard Deviation

Sampling distribution of x̄:

The distribution of values taken by x̄ from all possible sample of the same size from the same population.

Center of Sampling Distribution of x̄:

The mean of the sampling distribution of x̄ equals the population mean, μ.

Spread of Sampling Distribution of x̄:

The standard deviation of the sampling distribution of x̄ = σ/√(n)

Computing Probability Example 3: Closing prices of stocks have a right-skewed distribution with μ = 26 and σ = 20. What is the probability of randomly selecting a closing stock whose value is less than $15?

We can not calculate this using the z-score because the closing stock prices is not normal. If we did calculate it, the approximation would be poor.

Z-score for x̄:

Z = (x̄-value - mean of x̄) / (standard deviation of x̄) = (x̄ - μ) / (σ / √(n))