To Lecture Notes

IT 223 -- 3/7/11

Review Questions

1. When do you use a z-test?

   Ans: When you want to determine if an effect is significant, or just due to chance. Use the z-test if n > 30.

2. What are the five steps of a test of hypothesis?

   Ans:

   1. State the null and alternative hypotheses.
      
   2. Compute the test statistic, assuming that the null hypothesis is true.
      
   3. Write down the 95% or other confidence interval.
      
   4. If the test statistic is in the confidence interval, accept H₀, otherwise accept.
      
   5. Compute the p-value if this is a z-test, otherwise find it on the SPSS output.

3. Find the following confidence intervals for a z-test:
   65%    80%    98%    99.9%

   Ans: [-0.935, 0.935]    [-1.128, 1.128]    [-2.236, 2.236]    [-3.291, 3.291]

4. What is a p-value? Why is it important?

   Ans: The p-value is the probability that the test statistic is as extreme or more extreme than the value actually obtained, given that the null hypothesis is true.

5. Use the standard normal table to determine the p-value if z = 1.75.

6. 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.

   Ans: Here are the steps of the z-test:

   1. H₀ = 168      H₁ ≠ 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.

Tests of Hypothesis with Known or Unknown Independent Variables

• Reducing the effect of known independent variables:
  If the variable is known and can be controlled:
  -- Insure that the effect of the variable is the same on all subjects.

  If the variable is known, but cannot be controlled:
  -- Include the variable as an independent variable in the model (learn about this in a second statistics class).

  -- Include the variable as a panel variable (perform separate analyses for different values of the independent variable).

  -- Only include observations with the same value of the independent variable.

  If the values of the variable are unknown (is a lurking variable):
  -- Randomize the assignment of treatments to subjects to ensure that the various possible values of the lurking variables are equally likely for all subjects.

  -- If possible, make the treatments double blind.

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.

• SPSS never performs z-tests; use a t-test with SPSS, even if n ≥ 30.

• 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 statistic used in the t-test for μ is computed exactly the same way as the z-statistic for μ:
  t = (x - μ) / SE_ave,

  where μ is the value of μ assumed in the null hypothesis.

• The t-table is used instead of the z-table to account for the extra variability, due to the fact that x is not an accurate estimator of μ.

• Example 4:   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.

• The degrees of freedom for a t-test is related to the n - 1 that is used in the denominator of SD+.

• The degrees of freedom are the number of observations minus the number of parameters that must be estimated when estimating the variance for a statistical model.

• For example, when estimating the variance for the ideal measurement model
  σ² = [(x₁ - x)² + ... + (x_n - x)²] / (n - 1),
  we must estimate the parameter μ with x, so we loose one degree of freedom from the n degrees of freedom gained by the n observersions.

• When estimating the variance for a regression model, we have n - 2 degrees of freedom: +n for the n observations; -2 for estimating the two parameters a and b in y = ax + b.

t-tests with SPSS

• To perform a one-sample t-test with SPSS:
  Analyze >> Compare Means >> One-Sample T Test

• To perform a paired-sample t-test with SPSS:
  Analyze >> Compare Means >> Paired-Samples T Test

Practice Problems

• The datasets for these problems are found in the Excel file t-test1.xls.
  1. In 1998, as an adverting campaign, a cookie company claimed that every 18-ounce bag contained an an average of 1000 chocolate chips. Students at the Air Force Academy in Colorado Springs bought some randomly selected bags of cookies and counted the chocolate chips. Some of their data are recorded on the ChocolateChips dataset of the Excel file.
     1. Form the normal plot of the chocolate chip counts. Are the counts normally distributed?

     2. Perform a 99%-level test of hypothesis that the average chocolate chip count is 1000 per bag.

  2. Psychology experiments involve testing the ability of rate to navigate mazes. The mazes are classified according to difficulty, as measured by the average length of time it takes rats to find the food at the end. One researcher needs a maze that will take rats an average of about one minute to solve. He tests one maze on several rats, collecting the data on the RatMaze spreadsheet in the Excel file.
     1. Form the normal plot and box plot of the maze times. Are the times normally distributed?

     2. Test the hypothesis that the true completion time is one minute.

     3. Eliminate the outlier and perform the hypothesis test again.

  3. A food company marks the net weight of their potato chip bags as 28.3 grams. To test whether this claim is true, students collect and measure the net weights of bags. The measurements are recorded on the PotatoChipBags spreadsheet in the Excel file.
     1. Form the normal plot of the new weights of the potato chip bags. Are the weights normally distributed?

     2. Test the hypothesis that the true weight of potato chip bags is 28.3 grams.

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 2:   See the dataset t-test2.xls. 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:

  Type A	Type B
  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.

• Use SPSS to perform the paired-sample t-test:
  Analyze >> Compare Means >> Paired Sample T test.

Project 5

• Discuss Project 5.