Oct 26, 2016

IT 403 -- Oct 26, 2016

Review Exercises

Use the standard normal table to find these two-sided confidence intervals:
a. 99% b. 99.9 c. 80%
Ans: a. If the area under the normal curve over the interval [-z, z] is 99% = 0.99, then the areas corresponding to (-∞, -z] and [z, ∞) are both 0.005. This means that the area corresponding to (-∞, -z] ∪ [-z, z] = (-∞, z] is 0.99 + 0.005 = 0.995. Look up 0.995 in the body of the standard normal table and find the corresponding z-score: 2.575. Therefore the area corresponding to [-2.575, 2.575] is 0.99 = 99%.
b. The 99.9% two-sided confidence interval is [-3.29, 3.29].
c. The 80% two-sided confidence interval is [-1.28, 1.28].
Notes: one-sided confidence intervals of the form (∞, z] and [z, ∞) are also possible. We will discuss them later.
For a Bernoulli random variable x with probability of success p, what is the expected value E(X) and theoretical standard deviation σ_x?
Ans: For a Bernoulli random variable: E(x) = p and σ_x = √p(1 - p).
For a Binomial random variable x with n trials and probability of success p, what are E(S) and σ_S?
Ans: For a binomial random variable: E(S) = np and σ_S = √np(1 - p).
To forecast an election, an interviewer asks a random sample of people "Will you vote for Candidate X?" Here are the results:
n = 2,500 S = 1,376.
Find a 95% confidence interval for the true probability that a random person will vote for Candidate X. What is the sampling error?
What is a test of hypothesis? What are the null and alternative hypothesis?
Ans: A test of hypothesis is used to determine if there is a real difference between a random sample and the population. The null hypothesis, denoted by H₀, states that there is no real difference; the difference is just random variation. The alternative hypothesis, denoted by H₁, states that the difference is real.

Use SPSS to simulate the following situations. Use the random number generator Rv.Binom(n, p).

1,000 outcomes of a Bernoulli random variable with p = 0.5.
10,000 outcomes of a Bernoulli random variable with p = 0.7.
100,000 outcomes of a Bernoulli random variable with p = 0.9.

Ans: In SPSS, set up a dataset with two scale variables n and p like this:

n	p
1000	.5
10000	.7
100000	.9

Then perform these computations with Transform >> Compute Variable:

Target Variable	Numeric Expression	Description
S	RV.BINOM(n, p)	Generate binomial random outcomes.
SE_S	sqrt(n * p * (1 - p))	Compute standard error of the sum.
z	(S - n * p) / SE_S	Compute test statistic z.

Here are the results:

n	p	S	SE_S	z
1000	.5	498	15.81	-.1265
10000	.7	6949	43.83	-1.1129
100000	.9	90067	94.87	.7062

In the Variable View, the number of decimal places is set to 0, 1, 0, 2, and 4 for the variables n, p, S, SE_S, and z, respectively.

For each of the cases in Problem 6, test whether the random number generator is fair for the given probability.
Ans: The standard normal 95% two-sided confidence interval is I = [-1.96, 1.96]. In each case, z ∈ I, so we accept the null nypothesis that the binomial random number generator is fair in all cases.
If the probability of success is p, let x be the "time to failure" random variable, that is the number of successes of a Bernoulli random variable required to obtain a failure. This random variable x is called a geometric distribution.
1. What is the probability distribution of x?
  Ans: Set S = Success, F = Failure, x = number of successes before a failure:
  
  Outcome x P(X)
  
       F 0 1 - p
  
       SF 1 p(1 - p)
  
       SSF 2 p²(1 - p)
  
       SSSF 3 p³(1 - p)
  
        ... ...      ...
  
  To be a legitimate probability distribution, the sum of all the probabilities must be 1. To check, first let's find the sum of the geometric series:
        S = 1 + p + p² + p³ + ...        Equation 1.
  Multiply Equation 1 by p to obtain:
        pS = p + p² + p³ + p⁴ + ...     Equation 2.
  Now if we subtract Equation 1 minus Equation 2, the terms involving p go away and we obtain
        S - pS = 1.
  Solve for S to obtain S = 1 / (1 - p). Now we can sum the terms in the P(x) column of the probability distribution:
        1 - p + p(1 - p) + p²(1 - p) + p³(1 - p) + ...
  = (1 + p + p² + p³ + ... )(1 - p) = [1 / (1 - p)](1 - p) = 1.
2. What is E(x)? Ans:
  E(x) = 0P(0) + 1P(1) + 2P(2) + 3p(3) + ...
        = 0(1 - p) + 1p(1 - p) + 2p²(1 - p) + 3p³(1 - p) + 4p⁴(1 - p) + ...
        = p - p² + 2p² - 2p³ + 3p³ - 3p⁴ + 4p⁴ + ...
        = p + p² + p³ + p⁴ + ... )
        = p(1 + p + p² + p³ + ... ) = p[1 / (1 - p)] = p / (1 - p).

Outcome	x	P(X)
F	0	1 - p
SF	1	p(1 - p)
SSF	2	p²(1 - p)
SSSF	3	p³(1 - p)
...	...	...

Quiz 6

Go over Quiz 6 of the D2L Quizzes.

The Central Limit Theorem

The Central Limit Theorem (CLT) was first postulated by Abraham de Moivre in 1733 to estimate probabilities of the number of heads resulting from many tosses of a fair coin.
In 1901, the Russian mathematician Alexandr Lyapunov stated and proved the CLT in its modern form:
If x₁, ... , x_n are independent observations from the same random variable x of any distribution and n is large enough, then the sum
      S = x₁ + ... + x_n
is approximately normally distributed with expected value
      E(S) = nE(x₁)
and standard deviation
      SE_S = σ_x√n.
For our purposes, n is "large enough" if n ≥ 30.
The average
x = (x₁ + ... + x₁) / n
is also approximately normally distributed with expected value E(x) and SE_ave = σ_x / √n in this situation.

Practice Problems

Use the CLT to estimate the following probabilities for the number of heads obtained from a fair coin:
1. Obtaining 13 to 16 heads out of 25 tosses. (Because the normal table is continuous, use 12.5 to 16.5 tosses.)
  Ans: SE_S = √np(1-p) = √25(0.5)(1-0.5) = 2.5.
  Then z₁ = (S - np) / SE_S = (12.5 - 25(0.5)) / 2.5 = 0
  z₂ = (S - np) / SE_S = (16.5 - 25(0.5)) / 2.5 = 1.6;
  the area under the standard normal curve for the bin [0,1.6] is 0.9452 - 0.5000 - 0.4452 = 44.5%.
  The exact value obtained computing P(13) + P(14) + P(15) + P(16) is 0.4461 = 44.6%.
2. Obtaining between 60 to 75 heads out of 100 tosses. (Use 59.5 to 75.5 tosses.)
  Ans: SE_S = √np(1-p) = √100(0.5)(1-0.5) = 5.
  Then z₁ = (S - np) / SE_S = (59.5 - 100(0.5)) / 5 = 1.9,
  z₂ = (S - np) / SE_S = (75.5 - 100(0.5)) / 5 = 5.1;
  the area under the normal curve for the bin [1.9,5.1] is 1 - 0.9713 = 0.0287
  The exact value obtained using the binomial formula is 0.0284.
3. Obtaining exactly 30 heads out of 60 tosses. (Use 29.5 to 30.5 tosses.)
  Ans: SE_S = √np(1-p) = √60(0.5)(1-0.5) = 3.87.
  z₁ = (S - np) / SE_S = (29.5 - 60(0.5)) / 3.87 = -0.129,
  z₂ = (S - np) / SE_S = (30.5 - 60(0.5)) / 3.87 = 0.129;
  the area under the normal curve for the bin [-0.129, 0.129] is 0.5513 - 0.4487 = 0.1026 = 10%
  The exact value obtained using the binomial formula is 0.1026.

Tests of Hypothesis in General

The fundamental question that a researcher conducting a test of hypothesis is trying to answer is: Is the result significant, or is it merely due to chance variation?
I flip a coin 10,000 times and obtain 5,038 heads. Is my coin biased, or is it just chance variation?
Reducing the effect of lurking variables:
If the lurking variable is known and can be controlled:
-- Insure that the effect of the lurking variable is the same on all subjects.
If the lurking variable is known, but cannot be controlled:
      -- Include the lurking variable as an independent variable in the model.
      -- Include the lurking variable as a panel variable.
      -- Only include observations with the same value of the lurking variable.
If the lurking variable is unknown:
-- Randomize the assignment of treatments to subjects to ensure that the effect of lurking variables is equally likely for all subjects.
-- If possible, make the treatments double blind.
The null hypotheses, denoted by H₀, states that the treatment or effect under investigation does not make a difference; the effect is merely due to chance.
Some sample research questions phrased as null hypotheses:
1. The number of heads obtained with a coin being investigated is not significantly different than a fair coin.
2. There is no real difference in the autism rates between children that receive the vaccine and those that do not receive it.
3. The electric and magnetic fields caused by high voltage power lines does not cause significant health risks to those living nearby.
4. Eating irradiated food does not cause significant health risks.
5. There is no real difference in reading scores between the students that use the new reading curriculum and those that do not.
6. The faces the die are all equally likely to come up.
7. There is no real difference in network traffic speed between the old router and the new router.
8. The new tax law is essentially revenue neutral.
The steps for a test of hypothesis. The test is a α%-level z-test:
1. State the null (H₀) and alternative hypotheses (H₁).
2. Compute the test statistic T, assuming the null hypothesis is true.
3. Write down a (1-α)% confidence interval for the test statistic I.
4. If T ∈ I, accept the null hypothesis; if T ∉ I, reject the null hypothesis.
5. (If possible) determine the p-value of the test statistic.
Note: for Project 4, the z-tests are 5%-level tests.
The p-value is the probability of obtaining a test statistic at least as extreme at the test statistic actually obtained, given that the null hypothesis is true.

The z-test

Example 1: To test whether a new tax bill is revenue neutral, test the new tax rules on a random sample 100 tax returns out of 100,000 tax returns on file. For each tax return, compute
x_i = tax under new rules - tax under old rules
Then compute x = -$219 and SD = $725. Let μ be the true change in tax revenue from the old rules to the new rules.
The test of hypothesis:
1. State the null and alternative hypotheses:
  H₀: μ = 0
  H₁: μ ≠ 0
2. Compute the test statistic, assuming H₀ is true:
  z = (x - μ) / SE_ave = (-219 - 0) / (725 / √100) = -3.02
3. Write down a 95% confidence interval for the test statistic, assuming H₀: [-1.96, 1.96].
4. z = -3.02 ∉ (-1.96, 1.96), so reject H₀ and accept H₁; the difference is real and not merely due to chance.
5. Compute the p-value, which is the probability that z is as extreme or more extreme the z which is actually obtained. In our case z = -3.02, and we want to find the probability in the tails. Now the area under the normal curve for the interval (∞, -3.02) is 0.0013. Thus the area in both tails is 2 × 0.0013 = 0.0026.
The general form of the z-score for a z-test is
z = (T - E(T)) / SE_T
Two examples are
z = (S - E(S)) / SE_S and z = (x - E(x)) / SE_x
When performing a z-test for p, the test statistic is z = (S - np) / SE_S, where S is the number of successes from the Binomial Distribution (sum of n Bernouilli random numbers), and p is the true probability of success.
When performing a z-test for μ, from the ideal measurement model, the test statistic is
z = (x - μ) / SE_ave,
where μ is the mu of the null hypothesis.
Only use a z-test when the sample size n is greater than or equal to 30. This insures that, in the case of the ideal measurement model x_i = μ + e_i, the test statistic
z = (x-μ) / SE_ave
is approximately normally distributed according to the CLT, and
SD⁺ in SE_ave = SD⁺ / √n is close to the true standard deviation σ of the population.
In the case of a z-test for a probability p, n ≥ 30 insures that
the test statistic S is approximately normally distributed (thanks to the Central Limit Theorem), and
p^ = S/n in SE_S = √np(1-p) is close to the true value of p.

More about p-values

In the old days (50 years ago) statisticians were content to know whether H₀ was accepted or rejected. Now they want to know the p-value, which gives more information.
If p is close to zero, the evidence is overwhelming that the result was not due to chance.
If p is slightly less than 0.05, H₀ was just barely rejected. The evidence is borderline as to whether the result is due to chance.
If p is slightly more than 0.05, H₀ was just barely accepted. The evidence is borderline as to whether the result is due to chance; more research is required.
Usually a researcher wants to reject H₀ to prove that the treatment that he or she is investigating is real, not just chance variation.
If H₀ is accepted, it does not necessarily mean that we believe that H₀ is true, it means that there is not enough evidence to reject it.
At the risk of complicating things, our formulation of the z-test is phrased as a two-tailed test, where
H₀: μ = c H₁ ≠ c For a 5% level test (95% confidence), this means that we reject H₀ when the test statistic is not in the confidence interval I = [-1.96, 1.96].
It is also possible to phrase the z-test (and the t-tests we will discuss later) as one tailed tests, for which the null and alternative hypotheses are
H₀: μ = c H₁: μ > c
or
H₀: μ = c H₁: μ < c
In the case of H₁: μ > c, we reject H₀ when the test statistic is not in the interval I = (-∞, 1.645].
Some researchers think that the one-tailed test is an improvement over the two-tailed test because the test of hypothesis is more precise.
However, other researchers think that the one-tailed test is cheating, because it makes rejecting the null hypothesis easier.

Practice Problems

We will discuss the following two practice problems On November 3.

In 1999, it was reported that the mean serum cholesterol level for female undergraduates was 168 mg/dl. A recent study at Baylor university collected the following data for cholesterol levels for females:
x = 173.7 SD+ = 27
Is there a real difference between the women in the Baylor study and the reported value in 1999? (Example 6.15 from textbook). Perform the test at the 90%-level.
1. H₀: 168 H₁: μ ≠ 168
2. z = (x - μ) / SE_ave = (173.7 - 168) / (27 / √27) = 1.78
3. A 90% confidence interval for z is [-1.64,1.64].
4. 1.78 ∉ [-1.64,1.64], so reject the null hypothesis.
5. Find the area corresponding to the bin [-1.78,1.78]: 2 × 0.0375 = 0.0750.
Claim: if all high school seniors in California took the SAT test, the mean score would be equal to 450. To test this claim, take a sample of 400 high school seniors and give them the test. Here are the data:
n = 400 x = 461 SD+ = 100
Is this result for the sample significantly different from 450 or is it just chance variation? Perform the test at the 99%-level.
Ans: Here are the steps of the z-test:
1. H₀: μ = 450 H₁: μ ≠ 450
2. z = (x - μ) / SE_ave = (461 - 450) / (100 / √400) = 2.2.
3. A 99% confidence interval for z is [-2.58,2.58].
4. 2.2 ∉ [-2.58, 2.58], so reject the null hypothesis.
5. The p-value is the probability of obtaining a z-value as extreme or more extreme than the one actually obtained. Find the area corresponding to the bin [-2.2,2.2]: 2 × 0.0139 = 0.0278.

The t-test

Use the t-test for μ when n < 30 and the data are close to normally distributed.
Do not use the t-test for a sum.
Construct a normal plot to check if the data are close to normal whenever a t-test is performed.
Because n is small, SD might not be a good approximation of σ This increases the variability, which increases the size of the confidence interval.
The t-table is used instead of the z-table to account for the extra variability.
Example 2: A technician makes five measurements of the concentration of carbon monoxide (CO):
78 83 68 72 88
The descriptive statistics are:
n = 5 x = 77.8 SD⁺ = 8.07
Is the average of these concentrations significantly different than 70 or is it just chance variation?
There are two important differences between the t-test and the z-test:
1. The t-table is used instead of the z-table.
2. The p-value is hard to compute so we let SPSS compute it.
The form of the t-statistic is exactly the same as the z-statistic. The only difference is that since n <, we can no longer guarentee that t is normally distributed, so we use the t-table instead of the standard normal table.
Here is the t-test for Example 2.
1. State the null and alternative hypotheses:
  H₀: μ = 70
  H₁: μ ≠ 70
2. Compute the test statistic:
  t = (x - μ) / (SE / √n) = (77.8 - 70) / (8.07 / √5) = 2.16
3. Look up a 95% confidence interval using the t-table with n - 1 = 4 degrees of freedom. Use the upper-tail probility of 0.025 to obtain a two sided confidence interval of 95% to obtain 2.776.
4. Decide whether to accept or reject the null hypothesis:
  2.16 ∈ [-2.776, 2.776], so accept H₀.
To perform this test using SPSS, create a dataset with variable x containing 78, 83, 68, 72, 88. Then use
Analyze >> Compare Means >> One Sample t-test. Move x into the Test Variables box. Set the test variable to μ = 70, click Options and set the confidence level to 95%. Click OK.
A p-value of 0.097 is obtained, which means accept the null hypotheses, since p > 0.05.

Degrees of Freedom

Degrees of Freedom (df) is a technical term that arises when using the t-test. We are using x to estimate μ in SD+. If we are computing the SD+, when we are computing the square of the deviations, once we know the first n - 1 deviations, we automatically know the nth deviation because the sum of the deviations is always zero. n - 1 is called the degrees of freedom because only n - 1 of the deviations are able to vary freely.
The degrees of freedom for the t-test is related to the n - 1 that is used in the denominator of SD+.
Taking df = n - 1 compensates for the additional variation introduced because the true mean μ is unknown and x is used to estimate it.

The Paired Sample t-test

Goal: to test whether there is a significant difference between subjects from two different groups.
Typically, one group is the treatment group and the other group is the control group.
To use the paired sample t-test, each subject in one group is matched with a subject in the other group.
Then compute the differences in the response variable and perform a one-sample t-test on the differences.

Example 3: To test whether a new type of shoe sole material (type B) is better than the old type (type A), manufacture 10 pair of shoes where one shoe is made of type A and the other of type B. Randomly assign the type of material to left or right. Here is the data:

SoleMaterialA	SoleMaterialB
13.2	14.0
8.2	8.8
10.9	11.2
14.3	14.2
10.7	11.8
6.6	6.4
9.5	9.8
10.8	11.3
8.8	9.3
13.3	13.6

Perform the paired-sample t-test to see if there is a real difference between the two sole materials, or if it is just chance variation.
Into SPSS, import the Excel file t-test2.xlsx. Use the Paired Sample t-test Sheet.
Use SPSS to perform the paired-sample t-test:
Analyze >> Compare Means >> Paired Sample T test.
Here are the five steps of the two-sample t-test:
1. Write down the null and alternative hypothesis:
  H₀: SoleMaterialA = SoleMaterialB
  H₁: SoleMaterialA ≠ SoleMaterialB
2. Obtain the test statistic from SPSS: t = -3.349
3. Using the t-table, obtain a 95% confidence interval with n - 1 = 10 - 1 = 9 degrees of freedom:
  I = [-2.26, 2.26]
4. t ∉ I so reject H₀.
5. Find the p-value from the SPSS output (labeled Sig. (2-tailed)):
  p = 0.009.
The test statistic for the two-sample t-test obtained by computing the differences
diff = SolematerialA - SoleMaterialB
using Transform >> Compute variable in SPSS, and then performing a one-sample t-test on the variable diff.
Here are the five steps of the one-sample t-test performed with the diff variable:
1. Write down the null and alternative hypothesis:
  H₀: diff = 0
  H₁: diff ≠ 0
2. Obtain the test statistic from SPSS: t = -3.349
3. Using the t-table, obtain a 95% confidence interval with n - 1 = 10 - 1 = 9 degrees of freedom:
  I = [-2.26, 2.26]
4. t ∉ I so reject H₀.
5. Find the p-value from the SPSS output (labeled Sig. (2-tailed)):
  p = 0.009.
Notice that the test statistic t and the p-value are exactly the same whether a paired two-sample t-test is performed or whether a one-sample t-test on the differences is performed.