To Notes

IT 403 -- Nov 9, 2016

Review Exercises

1. Use the t-table to find a 95% confidence interval of the true CO level from the CO dataset:
         78  83  68  72  88
   Ans: n = 5  xbar = 77.8  SD+ = 8.07   SE_ave = 8.07 / sqrt(5) = 3.60
   From the t-table with 5 - 1 = 4 degrees of freedom, a 95% confidence interval is [-2.77, 2.77].
   Then about 95% of the time,
   -2.77 ≤ t ≤ 2.77
   -2.77 ≤ (xbar - μ) / SE_ave ≤ 2.77
   -2.77 ≤ (77.8 - μ) / 3.60 ≤ 2.77
   -9.972 ≤ 77.8 - μ ≤ 9.972
   -87.77 ≤ -μ ≤ -67.63
   67.6 ≥ μ 87.8
   Hence the 95% confidence interval is [67.6, 87.8].
2. Explain the differences and similarities between a one-sample z-test and a one-sample t-test. Ans:
   

   Test	Sample Size	Distribution
   z	> 30	Arbitrary
   t	Arbitrary	Normal

3. How do you decide whether to accept or reject the null hypothesis for a z- or t-test? Ans:
   
   1. Check if T is the test statistic and I is a confidence interval for this test statistic, accept H₀ if T ∈ I. Otherwise reject H₀.
   2. If α is the level of the test, accept H₀ if the p-value ≥ α,
      otherwise reject H₀.
   3. If I is a confidence interval for the true value of μ, accept H₀ if the null hypothesis value of μ is in I.
4. What is the definition of the p-value?
   Ans: The p-value is the probability of obtaining a test statistic as extreme or more extreme than the test statistic value actually obtained, given that the null hypothesis is true.
5. Find the 47th percentile of the following list using linear interpolation.
   Ans: See the bottom of the Know How To section of the Final Exam Review Guide -- compute a percentile of a dataset using the SPSS HAVERAGE method.

The Independent Two-sample t-test

• Here are the five steps of the independent two-sample t-test:
  Suppose that the two groups are labeled A and B:
  
  1. Write down the null and alternative hypothesis:
           H₀: μ_A = μ_B
  2. Compute the test statistic:
     
     
     
     where
     
     
     
  3. Find a 100(1 - α)% confidence interval I for t. Use n_A + n_B - 2 degrees of freedom.
  4. If t ∈ I, accept the null hypothesis H₀; if t ∉ I, reject the null hypothesis and accept the alternative hypothesis H₁.
  5. Use SPSS to obtain the p-value.
• Let's to back to the Calcium and Blook Pressure Example from last week to check how the calculations are performed.
• Example 1:   Calcium and Blood Pressure Example (Textbook Example 7.19) Does increasing the amount of calcium in diet reduce blood pressure? Use SPSS to perform an independent-sample t-test to check whether there is significant difference between the blood pressure decrease for the Calcium group and the Placebo group. Here is an Excel file with the data: calcium-blood-pressure.xslx. We perform the test first with SPSS. Later will check the results by hand.
  To perform the independent two-sample t-test with SPSS, do the following:
        Import the CalciumBloodPressure dataset into SPSS.
        Set the Value Label to Calcium for Group 0 and to Placebo for Group 1.
        Select Analyze >> Compare Means >> Independent Samples T-Test
        In the Independent Samples T-Test Dialog, drag Decrease into the Test Variables box. Also, drag Group into the Grouping Variables box.
        Click the Define Groups box. Enter 0 for Group 1; enter 1 for Group 2. Click Continue.
        Click OK.
  Here are the results. Look in the Independent Samples T-Test results in the first row: equal variances assumed:
        t = 1.634    df = 21    Sig. (2-tailed) = 0.096.
  This means that the p-value is 0.096: we reject the null hypothesis if we are testing at the 5%-level.
• Now let's check our results by hand, recalling that the groups are numbered 0 and 1:
  
  1.   H₀: μ₀ = μ₁
       H₁: μ₀ ≠ μ₁
  2.    n₀ = 10    x₀ =  5.00      SD₀+ = 8.743
        n₁ = 11    x₁ = -0.273    SD₁+ = 5.901
        SD_pooled² = [(n₀ - 1)s₀² + (n₁ - 1)s₁²] / (n₀ + n₁ - 2)
                     = [(10 - 1)8.743 + (11 - 1)5.901] / (10 + 11 - 2) = 54.536
         SD_pooled = √54.536 = 7.385
                   t = (x₁ - x₀) / [SE_pooled sqrt(1 / n₀ + 1 / n₁)]
                     = [5.000 - (-0.273)] / [7.285 sqrt(1/10 + 1/11)]
                     = 5.273 / (7.385 · 0.437) = 1.634
  3. The degrees of freedom are df = n₀ + n₁ - 2 = 10 + 11 - 2 = 19. Find a 95% confidence interval (0.025 upper tail area) from the t-table: I = [-1.729, 1.729].
  4. t = 1.634 ∉ [0, 1.729], so we accept H₀: we do not have enough evidence to reject the null hypothesis.

Chi-squared Distribution

• Let z be a random variable with the standard normal distribution.
• We write z ∼ N(0, 1) for short to mean z has a normal distribution with μ = 0 and σ² = 1.
• The random variable z² has a chi-squared (χ²) distribution with one degree of freedom (z² ∼ χ²(1)).
• Use SPSS to simulate 10,000 outputs from x1 = z² where z ∼ N(0, 1). We also know that x1 ∼ χ²(1):
        Run this script to set up the index variable i with values from 1 to 10,000
        See the document Enter a Range into a Dataset for details.
        Use Transform >> Compute Variable with
        Target Variable = x1; Numeric Expression: RV.NORMAL(0, 1)**2
        Now create a histogram and normal plot of the column x1.
• Use SPSS to simulate 10,000 outputs from x8 ∼ χ²(8):
        Run this script to set up the index variable i with values from 1 to 10,000
        See the document Enter a Range into a Dataset for details.
        Use Transform >> Compute Variable with
        Target Variable = x8;
        Numeric Expression: RV.NORMAL(0, 1)**2 + RV.NORMAL(0, 1)**2 +
            RV.NORMAL(0, 1)**2 + RV.NORMAL(0, 1)**2 +
            RV.NORMAL(0, 1)**2 + RV.NORMAL(0, 1)**2 +
            RV.NORMAL(0, 1)**2 + RV.NORMAL(0, 1)**2
        Now create a histogram and normal plot of the column x8.
        Notice that x8 is closer to a normal distribution than x1 is, thanks to the CLT.
  A shorter numeric expression to create χ²(8) outputs is
        RV.CHISQ(8)

Chi-squared Test for Goodness-of-fit

• Example 2:  Last week in class, we wanted to test the null hypothesis that brains of the class members are "fair" for selecting integers from 1 to 6 to simulate random die rolls. Each class member selected two random numbers between 1 and 6. Here are the numbers that were selected:
       4 1 1 1 4 2 2 4 2 6 2 3 2 6 6 2 2 6 1 3 3 4 3 5
  To test whether these numbers are a "fair" simulation of random die rolls, we will perform a chi-squared goodness-of-fit test.  Here is a table that summarizes the numbers:
  

  Outcome:	1	2	3	4	5	6
  Observed Count:	4	7	4	4	1	4
  Expected Count:	4	4	4	4	4	4

  
  Let O_i = observed count for outcome i and let E_i = expected count for outcome i. Here are the steps of the chi-squared goodness-of-fit test:
  1. If E_i is the expected count for Outcome i,
     H₀: E₁ = E₂ = ... = E₆,
     H₁: E_i ≠ E_j for some i and j.
  2. Compute the test statistic:
     X² = Σ (O_i - E_i)² / E_i
         = (4 - 4)² / 4 + (4 - 4)² / 4 + (7 - 4)² / 4 + (4 - 4)² / 4 + (1 - 4)² / 4 + (4 - 4)² / 4
         = 3² / 4 + (-3)² / 4 = 4.5
  3. The degrees of freedom = number of cells - 1 = 6 - 1 = 5. To find a 95% confidence interval, use the chi-squared table. Look in the 0.05 column and the 5 degrees of freedom row. The probability is 11.07 so the confidence interval is I = [0, 11.7].  (The confidence interval starts at 0 because the X² must always be nonnegative.
  4. X² = 4.5 ∈ I, so we accept the null hypothesis. We don't have enough evidence to conclude that our "human brain dice simulation" process is not fair.
  5. We will use SPSS to find the p-value: 0.480.
• To perform the chi-squared goodness-of-fit test with SPSS,
        Import the numbers from the Excel file human-dice.xslx.
        Set the Measure of the variable Number to Nominal.
        Select Analyze >>  Nonparameteric Tests >> Legacy Dialogs >> Chi-square ...
        Drag Number into the Test Variable List Box.
        In the Expected Value Box, leave the All categories equal radio button checked.
        Click OK.
  Here are the results:
        Test Statistic:            Chi-squared 4.5
        Degrees of Freedom:   df               5
        p-value:                    Asymp. Sig.  0.480
  Since p > 0.05, reject the null hypothesis that the die is fair.
• Example 3. (Textbook Example 9.15, the Adequate Calcium Today Study) In a study that explored the relationship between bone density and calcium intake, the investigators wanted to test whether the samples were representative of the participant pool, which consisted of 14,000 adolescents from six states Arizona (AZ), California (CA), Hawaii (HI), Indiana (IN), Nevada (NV), Ohio (OH):
  
  Number of study participants in the sample:
  

  AZ	CA	HI	IN	NV	OH	Total
  O1	O2	O3	O4	O5	O6
  167	257	257	297	107	482	1,567

  O_i = observed count for cell i = observed count of study participants in the sample.
  
  Proportions of participant pool from each state:
  

  AZ	CA	HI	IN	NV	OH	Total
  0.105	0.172	0.164	0.188	0.070	0.301	1.000

• Multiply the total number of study participants in the sample by the proportions of participants from each state to obtain the expected number of participants from each state. For example, the proportion of the participant pool from AZ is 0.105 and there are 1,567 total participants, so
  0.105 × 1,567 = 164.535.
  Expected count of participants from each state:
  

  AZ	CA	HI	IN	NV	OH	Total
  E1	E2	E3	E4	E5	E6
  164.54	269.52	256.99	294.60	109.69	471.67	1567.01

  E_i = expected count for cell i = expected count of participants for each state.
• We do not expect the expected counts to be exactly equal to the number of study participants from each state. The question is: is there a real difference between the numbers of study participants and the expected counts, or are the differences just random sampling errors?
• Perform a chi-squared goodness-of-fit test to see if there is a real difference between the observed counts O_i and the expected counts E_i:
  
  1. Write down the null and alternative hypotheses:
           H₀: Observed counts = Expected counts
           H₁: O_i ≠ E_i for at least one i.
  2. Compute the test statistic:
           χ² = Σ (O_i - E_i)² / E_i
               = (167 - 164.54)² / 164.54 + (257 - 269.52)² / 269.52 +
                  (257 - 256.99)² / 256.99 + (297 - 294.60)² / 294.60 +
                  (107 - 109.69)² / 109.69 + (482 - 461.67)² / 461.67
               = 0.930
  3. The number of degrees of freedom = the number of cells - 1 = 6 - 1 = 5.
     Use the chi-squared table to find a 95% confidence interval for df = 5.
     The confidence interval is I = [0, 11.07].
  4. Because 0.930 ∈ I, we accept H₀; there is no real difference between the observed cell counts and the expected cell counts.
• We can use SPSS to perform this chi-squared test.
  Import the Excel dataset adequate-calcium-today.xslx. State and StateNum should be set to Measure = Nominal, Observed and Proportion should be set to Scale:
  

  State	StateNum	Observed	Proportion
  AZ	1	167	0.105
  CA	2	257	0.172
  HI	3	257	0.164
  IN	4	297	0.188
  NV	5	107	0.070
  OH	6	482	0.301

  
  Use Transform >> Compute Variable to add a new column Expected defined by Proportion * 1567.
  
  Use Data >> Weight Cases. Select the Weight cases by radio button and use Observed as the Frequency Variable.
  
  To perform the chi-squared test, select
        Analyze >>  Nonparameteric Tests >> Legacy Dialogs >> Chi-square ...
        Drag StateNum into the Test Variable List Box.
        In the Expected Values Box, click the Values radio button, then enter by hand, the expected values one at a time:
        164.54  269.52  256.99  294.60  109.69  471.67
        Click OK.
  
  In the output, we see that the test statistic is computed as 0.930 and the p-value, denoted as Asymp. Sig., is 0.968.
• Example 5: You can use SPSS to simulate 60,000 dice rolls:
  
  • Use a script to create an empty dataset with 60,000 rolls, indexed by i.
    Generate dice rolls in the roll variable with the numeric expression:
          TRUNC(RV.UNIFORM(0, 6)) + 1
    Perform a chi-squared goodness-of-fit test like you did in Example 4.
    A sample run produced df = 5, χ² = 6.455, p = 0.264

Chi-squared Test for Independence

• Example 6: (Textbook Example 9.1, Are you spending more time on Facebook?) Here is a two-way table that classifies Facebook users according to gender and whether or not the amount of time that they spent on facebook has increased:
  

  	Gender
  HasIncreased	Women	Men	Total
  Yes	47	21	68
  No	245	212	457
  Total	292	233	525

  
  We want to test whether gender and whether the subject's time on Facebook has increased.
  
  
  1. Compute the expected cell counts, assuming that HasIncreased and Gender are independent.
     According to the marginal totals, P(Yes) = 68 / 525 = 0.1295, P(No) = 1 - 0.1295 = 0.8705.
     Also, P(Woman) = 292 / 525 = 0.5562, P(Man) = 1 - 0.5562 = 0.4438.
     Therefore, if HasIncreased and Gender are independent,  P(Yes and Woman) = P(Yes)P(Woman) = 0.1295 × 0.5562 = 0.0720.
     Similarly, the probabilities for the other cells are P(No and Woman) = 0.4842, P(Yes and Man) = 0.0575, P(No and Man) = 0.3863.
     We can then multiply the grand total by each of these proportions to obtain the expected counts:
           Count for HasIncreased = Yes,  Gender = Woman: 525 × 0.0720 = 37.8
           Count for HasIncreased = No,   Gender = Woman: 525 × 0.4842 = 254.2
           Count for HasIncreased = Yes,  Gender = Man: 525 × 0.0575 = 30.2
           Count for HasIncreased = No,   Gender = Man: 525 × 0.3863 = 202.8
     These expected counts are summarized in this table:
     

     	Gender
     HasIncreased	Women	Men	Total
     Yes	37.8	30.2	68
     No	254.2	202.8	457
     Total	292	233	525

     
     
  2. Compute the test statistic:
           χ² = Σ (O_i - E_i)² / E_i
               = (47 - 37.8)² / 37.8      + (21 - 30.2)² / 30.2 +
                  (245 - 254.2)² / 254.2 + (212 - 202.8)² / 202.8
               = 5.79
  3. The number of degrees of freedom = (number of rows - 1) × (number of columns - 1) = 1 × 1 = 1.
     Use the chi-squared table to find a 95% confidence interval for df = 1.
     The confidence interval is I = [0, 3.84].
  4. Because 5.79 ∉ I, we reject H₀ that HasIncreased and Sex are independent.
  5. The p-value is computed from SPSS. However, we can get a rough idea of the p-value by looking at the t-table; since 5.41 < 5.79 < 6.63, we conclude that  0.02 > p > 0.01.
• To obtain the test statistic and the p-value from SPSS, perform the following steps:
  
  1. In an SPSS dataset, create these variables:
     

     Name	Type	Label	Values	Measure
     HasIncreased	Numeric	Has Facebook Usage Increased?	1="Yes",2="No"	Nominal
     Gender	Numeric	Gender of Subject	1="Male",2="Female"	Nominal
     Frequency	Numeric	Weighting Factor		Scale

     
     and enter the following data:
     

     HasIncreased	Gender	Frequency
     1	1	47
     2	1	245
     1	2	21
     2	2	212

  2. Enter these variable value labels:
     

     Variable	Value Format
     HasIncreased	{1 = "Yes", 2 = "No"}
     Gender	{1 = "Female", 2 = "Male"}

  3. Set up Frequency as the weighting factor for the dataset:
           Select Data >> Weight Cases ...
           In the Weight Cases Dialog, check the Weight cases by radio button.
           Drag Frequency into the Frequency Variable box; click OK.
  4. To perform the chi-squared test:
           Select Analyze >> Descriptive Statistics >> Crosstabs ...
           Drag Gender into the Column(s) box; drag HasIncreased into the Row(s) box.
           Click the Statistics Button. Check the Chi-squared box. Click Continue.
           Click OK.
  5. From the SPSS output, we obtain the following:
           Chi-squared value = 5.766;   df = 1;   p-value (Asymp. Significance) = 0.016
     We reject the null hypothesis that HasIncreased and Gender are independent: whether or not FaceTime usage has increased is not independent of gender.
• Practice Problem: (Textbook Example 9.8) Are Fruit Consumption and Physical Activity independent?
  Import the Excel Dataset: physical-activity-fruit.xslx
  

  	PhysicalActivity
  FruitConsumption	Low	Moderate	Vigorous	Total
  Low	698	206	294	569
  Medium	25	126	170	321
  High	14	111	169	294
  Total	108	443	633	1,184

Simpsons Paradox

• Gender Bias in Berkeley Grad School Admissions

Bootstrapping

• Look at this introduction to Bootstrapping.

Misuses of Statistics

• Look at the document Misuses of Statistics.
• You might find this paper interesting:
        The Human Cost of War in Iraq.
  Do you think that this paper does a good job of presenting statistics and using statistics to make its case?

The Power of a Statistical Test

• See Page 402 and following in the Moore, McCabe, and Craig textbook.
• For a specific value of the alternative hypothesis, the power of the test is the probability of rejecting the null hypothesis.
• High power is desirable. Along with 95% confidence intervals and 5% level tests, tests with 80% power have become the standard.
• Many U.S. government agencies that provide research funds require tests of hypothesis to have at least 80% power.
• Example 4: Can a six-month exercise program increate the total body bone mineral content (TBBMC) of young women?
  Let x be the percentage increase in TBBMC over the six month exercise period.  Here is what we know:
        n = 25   x = SD_x = 2    H₀: μ = 0
  What is the power of the test for the alternative H₁: μ = 1?
  Ans: First let's find the value of x needed reject H₀. (-∞, 1.645] is a one-sided 95% confidence interval for the test statistic x.
  H₀ is rejected when z = (x - μ₀) / (SD_x / √n) = (x - 0) / (2 / √25) = x / 0.4 ≥ 1.645.
  That is, when x / 0.4 ≥ 1.645 or x ≥ 0.658.
  
  Now, what is P(x ≥ 0.658 when μ = 1)? It is
        P((x - μ) / (σ/√n) ≥ (0.658 - 1) / (2 / √25))
                           P(z ≥ -0.855) = 0.80.
  Thus the probability of rejecting H₀ for μ = 1 is 0.80: the power of the the test when μ = 1 is 0.80. Here is a graph if the power of this test vs. the alternative hypothesis value.