1 / 33

Hypothesis Testing: Basic Concepts and Tests of Association, Chi-Square Tests

Hypothesis Testing: Basic Concepts and Tests of Association, Chi-Square Tests. Basic concepts - Example. GEICO feels that consumers do not like the caveman ad campaign so it needs to be changed

jerzy
Download Presentation

Hypothesis Testing: Basic Concepts and Tests of Association, Chi-Square Tests

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Hypothesis Testing: Basic Concepts and Tests of Association, Chi-Square Tests

  2. Basic concepts - Example • GEICO feels that consumers do not like the caveman ad campaign so it needs to be changed • GEICO wants to verify this feeling so they survey a sample and find that the campaign is well liked. • Should GEICO conclude that their feeling is wrong or that the sample mean is a function of chance?

  3. Hypothesis Testing: Basic Concepts • Hypothesis: An assumption made about a population parameter (not sample statistic) • E.g. Consumers dislike the caveman ad campaign • Purpose of Hypothesis Testing: To make a judgment about the difference between the sample statistic and the population parameter • The sample likes the caveman campaign. Is this an accurate representation of the population’s attitude? • The mechanism adopted to make this objective judgment is the core of hypothesis testing

  4. Hypothesis testing: Logic • Is the sample statistic a function of chance or luck rather than an accurate representation of the population parameter? • Example: • Hypothesized mean attitudes are 2 (on a 1(SD) – 5(SL) scale) • Observed mean attitudes are 4 (on a 1(SD) – 5(SL) scale) • Is the difference between the two a chance event or are we really wrong about our hypothesis? • This is statistically evaluated.

  5. Problem Definition Clearly state the null and alternative hypotheses. Choose the relevant test and the appropriate probability distribution Determine the degrees of freedom Determine the significance level Choose the critical value Compare test statistic and critical value Compute relevant test statistic Decide if one-or two-tailed test Does the test statistic fall in the critical region? No Do not reject null Yes Reject null

  6. 1. Formulate Null & Alternative hypotheses • Null hypothesis (Ho) – • the hypothesis of no difference • between the population parameter and sample statistic • OR no relationship • Between two population parameters • A mirror-image of the alternative (research) hypothesis • Alternative hypothesis (Ha or H1) – the hypothesis of differences or relationships in the population • Example • Ho: Mean population attitudes = 2 • Ha: Mean population attitudes are not = 2; OR • Ho: Use of social media is not related to likelihood of response to online ads • Ha: Use of social media is positively related to likelihood of response to online ads

  7. 2. Choose appropriate test and probability distribution • Depends on whether we are • Comparing means (Z distribution if population standard deviation is known; t distribution if population standard deviation is not known) • Comparing frequencies (chi-square distribution)

  8. 3. Determine significance level • The level at which we want to make a judgment about the population parameter (the null hypothesis) • Generally 10%, 5%, 1% (corresponding to 90%, 95% and 99% confidence levels) in social sciences • The level at which the critical test statistic is identified

  9. 4. Determine degrees of freedom • Number of bits of unconstrained data available to calculate a sample statistic • E.g. for X bar, d.f. is = n; for s, d.f. is n-1, since 1 d.f. is lost due to the restriction that we need to calculate the mean first to calculate the standard deviation

  10. 5. Decide if it is a one / two tailed test • One Tailed test: If the Research Hypothesis is expressed directionally: • E.g. Head-On wants to test if consumers dislike their ad campaign (mean liking < 3; (1 (strongly dislike) – 5 (strongly like) scale). • Ho: Population mean attitudes are greater than or equal to 3.0 • Ha: Population mean attitudes are less than 3.0 • For confirmation of Ha look in the tail of the direction of the Research Hypothesis

  11. 5. Decide if it is a one / two tailed test • Two Tailed test: If the Research Hypothesis is expressed without direction • E.g. Head-On wants to test if consumers feel differently about their ad campaign than they felt a year ago. (mean liking = 4.5; (1 (strongly dislike) – 5 (strongly like) scale). • Ho: Population mean attitudes = 4.5 • Ha: Population mean attitudes are not equal to 4.5 • For confirmation of Ha look in the tails on both sides of the distribution

  12. 6. Find the critical test statistic • Critical z value requires knowledge of level of significance • Critical t value requires knowledge of level of significance and degrees of freedom • Critical chi-square requires knowledge of level of significance and degrees of freedom

  13. 7. Criteria for rejecting / not rejecting H0 • Compute observed test statistic • Compare critical test statistic with observed test statistic • If the absolute value of observed test statistic is greater than the critical test statistic, reject Ho • If the absolute value of observed test statistic is smaller than the critical test statistic then Ho cannot be rejected. • Regions of rejection / acceptance

  14. Type 1 and Type 2 errors Data Analysis conclusion is: Null hypothesis in population is True False Reject Null hypothesis Do not reject Null hypothesis

  15. Type 1 and Type 2 errors • The lower the confidence level, the greater the risk of rejecting a true H0 – Type 1 error (alpha) i.e. you increase the chances of accepting a false research hypothesis • i.e. if you reduce the confidence level from 95% to 90% the chances of you declaring that the effect observed in the sample actually prevails in the population, are higher. • If the effect in reality does not exist in the population, then you commit a Type 1 error. • Therefore in Type 1 error you declare an effect which does not exist

  16. Type 1 and Type 2 errors • The higher the confidence level the greater the risk of accepting a false H0 – Type 2 error (beta), i.e. you reduce the chances of accepting a true research hypothesis • i.e. if you increase the confidence level from 95% to 99%, the chances that you miss the effect which may actually be there in the population, are higher. • the power of the test to spot the effect is reduced • Therefore power = 1 – beta • Therefore in Type 2 error you miss an effect which exists

  17. Hypothesis Testing Tests in this class Statistical Test • Frequency Distributions 2 • Means (one) z (if  is known) t (if  is unknown) • Means (two) t • Means (more than two) ANOVA

  18. Chi-Square as a test of independence • Statistical Independence: if knowledge of one does not influence the outcome of the other • E.g. Affiliation to school (nominally scaled) does not influence decision to eat at the student union • Expected Value: The average value in a cell if the sampling procedure is repeated many times • Observed Value: The value in the cell in one sampling procedure • Only nominal / categorical variables

  19. Chi-square Step-by-Step 1) Formulate Hypotheses

  20. Chi-Square As a Test of Independence Null Hypothesis Ho • Two (nominally scaled) variables are statistically independent • There is no relationship between school affiliation and decision to eat at the student union Alternative Hypothesis Ha • The two variables are not independent • School affiliation does influence the decision to eat at the student union

  21. Chi-square As a Test of Independence (Contd.) Chi-square Distribution • A probability distribution for categorical data • Total area under the curve is 1.0 • A different chi-square distribution is associated with different degrees of freedom

  22. F(x2) df = 4  = .05 x2 The chi-square distribution

  23. Chi-square Step-by-Step 1) Formulate Hypotheses 2) Calculate row and column totals 3) Calculate row and column proportions 4) Calculate expected frequencies (Ei) 5) Calculate 2 statistic

  24. Chi-square Statistic (2) • Measures of the difference between the actual numbers observed in cell i (Oi), and number expected (Ei) under independence if the null hypothesis were true With (r-1)*(c-1) degrees of freedom r = number of rows c = number of columns • Expected frequency in each cell: Ei = pc * pr * n Where pc and pr are proportions for independent variables and n is the total number of observations

  25. Chi-square Step-by-Step 1) Formulate Hypotheses 2) Calculate row and column totals 3) Calculate row and column proportions 4) Calculate expected frequencies (Ei) 5) Calculate 2 statistic 6) Calculate degrees of freedom

  26. Chi-square As a Test of Independence (Contd.) Degree of Freedom v = (r - 1) * (c - 1) r = number of rows in contingency table c = number of columns

  27. Chi-square Step-by-Step 1) Formulate Hypotheses 2) Calculate row and column totals 3) Calculate row and column proportions 4) Calculate expected frequencies (Ei) 5) Calculate 2 statistic 6) Calculate degrees of freedom 7) Obtain Critical Value from table

  28. The chi-square distribution F(x2) Critical value = 9.49 df = 4 • Ex: Significance level = .05 Degrees of freedom = 4 CVx2 = 9.49 5% of area under curve  = .05 x2

  29. Chi-square Step-by-Step 1) Formulate Hypotheses 2) Calculate row and column totals 3) Calculate row and column proportions 4) Calculate expected frequencies (Ei) 5) Calculate 2 statistic 6) Calculate degrees of freedom 7) Obtain Critical Value from table 8) Make decision regarding the Null-hypothesis

  30. Example of Chi-square as a Test of Independence Eat / Don’t eat YN A 10 8 School B 20 16 C 45 18 D 16 6 E 9 2 This is the observed value This is a ‘Cell’

  31. 0.24 * 0.67 * 150 36/150 Chi-square example

  32. Chi-square example • Observed chi-square = [(10 – 12)2 / 12] + [(8 – 6)2 / 6] + [(20 – 24)2 / 24] + …+ [(2 – 4)2 / 4] = 5.42 • d.f. = (r-1)(c-1) = (5-1)(2-1) = 4 • Critical chi-square at 5% level of significance at 4 degrees of freedom = 9.49 • Since observed chi-square < critical chi-square (5.42 < 9.49), H0 cannot be rejected • Hence decision to eat / not eat at the student union is statistically independent of their school affiliation. In other words there is no relationship between the decision to eat at the SU and the school they are in.

  33. The chi-square distribution F(x2) Critical value = 9.49 df = 4 Ex: Significance level = .05 Degrees of freedom = 4 CVx2 = 9.49 The decision rule when testing hypotheses by means of chi-square distribution is: If x2 is <=CVx2, accept H0 Thus, for 4 df and  = .05 If x2 is > CVx2, reject H0 If If x2 is <= 9.49, accept H0 5% of area under curve  = .05 x2

More Related