Unit 5 STATS
Center for one population
"Random Sample"
11) A simple random sample of size n = 64 is obtained from a population with 𝜇 = 50 and 𝜎 = 2. Does the population need to be normally distributed for the sampling distribution of 𝑥̅to be approximately normally distributed? (A) Yes, because the Central Limit Theorem states that the sampling variability of not normal populations will increase as the sample size increases. (B) No, because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of 𝑥̅becomes approximately normal as the sample size, n, increases. (C) No, because the Central Limit Theorem states that only if the shape of the underlying population is normal or uniform does the sampling distribution of 𝑥̅become approximately normal as the sample size, n, increases. (D) Yes, because the Central Limit Theorem states that only for underlying populations that are normal is the shape of the sampling distribution of 𝑥̅normal, regardless of the sample size n. (E) No, because our population mean is larger than 30, so the Central Limit Theorem allows us to say the sampling distribution will be approximately normal.
(B) No, because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of 𝑥̅becomes approximately normal as the sample size, n, increases.
Decreasing the sample size from 750 to 375 would multiply the standard deviation by (A) 2 (B) 1/2 (C) √2 (D) 1/√2 (E) None of these
(C) √2
Three steps in Evaluating a Claim
1. Assume the claim is true 2. Create a simulated sampling distribution 3. Find % chance of getting the observed result or more extreme
2 options to prove that a sampling distribution is a normal distribution
1. If the population distribution is normal, then so is the sampling distribution of x̄ 2. If the population is not normal or unknown, the sampling distribution of x̄, will be approximately normal in most cases if n is ≥ 30.
Variability for one population
10% (pseudo independence) n ≤ 0.10(n)
x̄ ~ norm (µx̄ = µ, = σx̄ = σ/square root of n What is square root of n
10% condition
When sampling we have to make sure the __ __ is satisfied. When we are not sampling, the problem must state that all "trails" are __
10% condition, independent
For explain why you can use this formula for the standard deviation? What do you put?
Because the 10% condition is satisficed
Center for two populations
Both "Random samples"
Shape for two populations
Both normal pop or CLT
Why is this description (Random Condition) important?
Creates an unbiased estimator so Mp̂ = p
A statistic is biased if the spread of the sampling distribution is not equal to the true value of the parameter. True or False?
False
Samples G and H were selected from the same population of quantitative data and the mean of each sample was determined. The mean of sample G is equal to the mean of the population. Which of the following statements must be true? The mean of sample H must also be equal to the population mean. The mean of sample G, is a point estimator for the mean of the population. The mean of sample H, is a point estimator for the mean of the population.
II AND III
For are you convinced that the E-Nose was more effective than random guessing? Why or why not? problem....... What do you put?
If the subject is guessing there is a __ answer__ % probability of guessing __ lower boundary value__ % correctly purely by chance. Because __ answer__% is __< or >__ than 5%, the __subject __is/is not__ likely to be guessing.
Parameter
Is a number describing a population
Statistic
Is a number describing a sample
How to find random mean? (formula)
Mp̂ = P = __
Random Condition description
Must have a random sample if sampling
Why is the Large Counts Condition description important?
Needs enough successes or failures to create a normal shape
x̄ ~ norm (µx̄ = µ, = σx̄ = σ/square root of n) What is norm?
Normal Population or CLT
The distribution of age for players of a certain professional sport is strongly skewed to the right with mean 26.8 years and standard deviation 4.2 years. Consider a random sample of 4 players and a different random sample of 50 players from the population. Which of the following statements is true about the sampling distributions of the sample mean ages for samples of size 4 and samples of size 50 ?
Only the sampling distribution for size 50 will be approximately normal, and the mean for both will be 26.8.
The continuous random variable NN has a normal distribution with mean 7.5 and standard deviation 2.5. For which of the following is the probability equal to 0 ?
P(N=8)
At a large corporation, 6,000 employees from department A and 4,000 employees from department B are attending a training session. A random sample of 500 employees attending the session will be selected. Consider two sampling methods: with replacement and without replacement. How will the methods affect the standard deviations of the sampling distribution of the sample proportion of employees from department B?
Sampling without replacement will result in a standard deviation less than but close to √0.4(0.6)/500.
Sampling Distribution
Shows the statistic found in all possible samples of size "n"
Why is this description (10% Condition) important?
Since we sample without replacement, we need a large enough population to approximate independence
A sample of 50 lions and their averages was 200 pounds with st deviation of 15 pounds. Turns out all mountain lions have a mean of 210 pounds with a st deviation of 25 pounds. __ = 15
Sx = 15
Interpret SD deviation response
The difference in sample proportions who have a __ typically varies by __standard deviation answer__ from the true difference of __mean answer__
Which of the following conditions will create a biased estimator of a population parameter?
The expected value of the estimator is not equal to the population parameter.
A certain statistic will be used as an unbiased estimator of a parameter. Let JJ represent the sampling distribution of the estimator for samples of size 40, and let KK represent the sampling distribution of the estimator for samples of size 100. Which of the following must be true about JJ and KK ?
The expected values of JJ and KK will be equal, and the variability of JJ will be greater than the variability of KK.
A manufacturing company uses two different machines, A and B, each of which produces a certain item part. The number of defective parts produced by each machine is about 1 percent. Suppose two independent random samples, each of size 100, are selected, where one is a sample of parts produced by machine A and the other is a sample of parts produced by machine B. Which of the following is true about the sampling distribution of the difference in the sample proportions of defective parts?
The mean is 0 and the distribution will not be approximately normal.
Which of the following would be the best reason why the simulation of the sampling distribution is not approximately normal if something is skewed.
The sample size was not sufficiently large.
What happens if a sample is higher than 30 but there is no information about mean to help with shape distribution?
The samples are sufficiently large (no information about normal distribution meaning we don't know if they are normal)
For calculate the probability of getting this difference in proportions or higher. how do you interpret?
There is a __percent answer__% of getting a difference of __ proportion given__ or higher in a sample of __total number + subject
For what is the probability that the proportion of red beads from Bin B is higher than proportion of Bin A questions.
There is a __percent found__ probability that the proportion of __ is higher than __ for samples of size __
The 10% condition makes sure our sample size is less than 10% of our population size. True or false?
True
Researchers are studying two populations of wild horses living in the western regions of a country. In a random sample of 32 horses taken from the first population, the mean age of the sample was 21 years. In a random sample of 41 horses from the second population, the mean age of the sample was 19 years. Is the sampling distribution of the difference in sample mean ages approximately normal?
Yes, because the sample sizes are both greater than 30.
If a sampling distribution is unbiased why is it so for (proportions)
because it has random guesses
Why should you not check for the 10% condition
because questions are independent
Proportions equal
categorical data
If the population is not normal or unknown, the sampling distribution of x̄, will be approximately normal in most cases if n is ≥ 30 is an example of
central limit theorem
when there is less variability, x̄ gets
closer together
When increasing sample size, the sampling distribution variability does what?
decreases
what makes a curve fit?
if something is approximately normal
simulated sampling distribution
is a distribution where we take ALL possible samples of a given size and put those sample statistics together as a data set
μ is sometimes __ known, but most of the time it is __
known, unknown
smaller standard deviation equals
less variation
μx
mean of all the sample means
A statistic is an unbiased estimator of the parameter if the __ of the __ __ is equal to the __
mean, sampling distribution, parameter
10% condition description
n ≤ 0.10(N)
Variability for two populations
n1 ≤ 0.10(N) n2 ≤ 0.10(N)
shape for one population
normal population or n ≥ 30 (CLT)
Large Counts Condition description
np ≥ 10 n(1-p) ≥ 10
Out of everyone who applies to UCLA, 58% are accepted. In a random sample of 100 high schoolers, 62% are accepted. __ = 0.58
p = 0.58
p̂ means
p-hat
10% of the population is left handed . In a random sample of 40 people, 12% are left handed. __ = 0.10
p=0.10
We use the statistic from the sample to estimate the
parameter
how to find shape (formula)
population number (percent that is one thing) = __ ≥ 10 and population number (1-percent that is one thing) = __ ≥ 10
σ represents the
population standard deviation
Out of everyone who lives in the metro area, 23 % are science majors. In a random sample of people in the metro area, 29% are science majors. __ = 0.29
p̂ = 0.29
In a random sample of 30 Americans 43% were female. __ = 0.43
p̂ = 0.43
Sampling distribution for a proportion formula (using a normal curve)
p̂ ~ Normal (μp̂ = p, σp̂ = square root of p(1-p)/n)
Means equal
quantitative data
Things to go over
questions on drawings she made
x̄ ~ norm (µx̄ = µ, = σx̄ = σ/square root of n What is µ?
random condition (Unbiased center)
x̄ represents the
sample mean
Sx represents the
sample standard deviation
A statistic is an unbiased estimator of the population when?
the mean of the sampling distribution equals the parameter
μ represents
the population mean
when something is random it also means that it is
unbiased
x̄ is an __ estimator of μ
unbiased
As long as the sampling method is random, our mean is an
unbiased estimator
When there is less variability n goes __ and s goes __
up, down
If the sample mean is an unbiased estimator of the population we can
use x̄ to estimate μ
x̄ means
x-bar
the most common statistical measure
x̄
In a sample of 20 professional players the average speed of their pitch was 55 mph with standard deviation of 3 mph. __ = 55
x̄ = 55
mean, proportion, and standard deviation sign for statistic
x̄, p̂, Sx
Mean formula
µp̂ = p
The national high school average sat score is 1060 with st deviation of 217. A random sample of 50 students who took the SAT were collected and the average score was 1100 with st deviation of 200. __ = 1060
μ = 1060
mean, proportion, and standard deviation sign for parameter
μ, P, σ
Out of all of the results at a local university, the average GPA is 2.8 with standard deviation of 0.65. A random sample of 30 students were selected and their average GPA was 3.1 with standard deviation of 0.2. __ = 2.8
μ=2.8
Out of all the students at a school, the average GPA was 2.8 with standard deviation of 0.65. A random sample of 30 students were selected and their averages was 3.1 with standard deviation of 0.2. __ = 0.65
σ = 0.65
A sample of 50 lions and their averages was 200 pounds with st deviation of 15 pounds. Turns out all mountain lions have a mean of 210 pounds with a st deviation of 25 pounds. __ = 25
σ = 25
The average height of basketball players is 78 inches with standard deviation of 3.5 inches. __ = 3.5
σ= 3.5
how to find 10% condition Standard Deviation? (formula)
σp̂ = square root of (percent something does happen) (percent does not happen) divided by 50
Standard Deviation formula
σp̂ = square root of p(1-p)/n
