To Lecture Notes

IT 223 -- 3/2/11

Review Questions

1. Flip a fair coin 1,000 times. Use the CLT to estimate the the probability of getting more than 510 heads? (Use S = 509.5)

   Ans: n = 1000, S = 509.5, p = 1/2, SEs = sqrt(n * p * (1-p)) = sqrt(1000 * (1/2) * (1-1/2)) = 15.8. Then

   z = (S - E(S)) / SEs = (S - np) / SEs = (509.5 - 1000*(1/6)) / 15.8 = 0.601.

   Look up -z = -0.601 in the standard normal table to get 0.2743 = 27%.

2. Use the CLT to estimate the probability that the rain on the tropical island in one year is between 400 and 410 inches. Recall that the expected value of rain in one day is 1.1 inches, with SD = 0.943 inches. (Use 399.5 to 410.5 inches).

   Ans: Recall that E(x) = 1.1, σ_x = 0.943, so E(S) = nE(x) = 365(1.1) = 401.5, σ_S = σ_x√n = 0.943√365 = 18.0.
   Then z₁ = (399.5 - 401.5) / 18.0 = -0.11 and z₂ = (410.5 - 401.5) / 18.0 = 0.5.
   The area corresponding to the bin [-0.11,0.5] under the standard normal curve is 0.6915 - 0.4558 = 0.2357 = 24%.

3. What is a null hypothesis?

   Ans: It is the assumption that unknown model parameter is equal to a specified value. For example, when testing whether a coin is fair, the null hypothesis is that p = 1/2.

4. What does it mean to reject a null hypothesis?

   Ans: It means that the value of the test statistic is not found in the confidence interval. The difference is due to chance, and is not chance variation.

5. What does it mean to accept a null hypothesis?

   Ans: It means that the value of the test statistic is found in the confidence interval. The difference is due to chance: just chance variation.

Tests of Hypothesis

• 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 5038 heads. Is my coin biased, or is it just chance variation?

• 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 given coin 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 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).

• 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 according to the CLT, 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.

Practice Problems

1. 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.

2. 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.

The z-test for a Sum

• Example 2:   Test whether the SPSS Bernoulli random number generator is fair for p = 0.5. The data obtained with n = 400 is S = 209. Perform the test at the 98%-level.
  1. State the null hypothesis and the alternative hypothesis:
     H₀: p = 0.5
     H₁: p ≠ 0.5

  2. Compute the test statistic:
     z = (S - np) / SE_S = (S - np) / √np(1-p)

         = (209 - 400(0.5)) / √400(0.5)(1-0.5) = 0.9

  3. Write down a 98% confidence interval:
     (-2.33, 2.33)

  4. Decide whether to accept or reject the null hypothesis:
     0.9 ∈ (-1.96, 1.96)

     so accept H₀; there is not enough evidence to conclude that the SPSS Bernoulli random number generator is not fair for p = 0.5.

  5. Compute the p-value. Compute the area under the normal curve for the bins (∞, 0.9] and [0.9, ∞), which is the two-sided tail probability. This gives us p = 0.2681.

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 3:   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 < 30, 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 3.
  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 x has been computed, and the data values x₁, ... , x_n-1 have been collected, then x_n is completely determined by the equation
  (x₁ + ... + x_n-1) / n = x.

• 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 SD σ is unknown and SD+ is used to estimate it.

Misuses of Statistics

• Look at the document Misuses of Statistics.