Workshop Statistics: Discovery with Data, Second Edition

Topic 26: Inference for Two-Way Tables

Activity 26-1: Suitability for Politics (cont.)

(a) observational units = sample of American adults
variable 1: whether or not they agree with the statement; categorical binary
variable 2: their political view: categorical
(b) Tend to have more agreement with statement as lean more conservative. While there is definitely a trend it may not be all that strong.
(c) agreement proportion = 382/1685 = .227
(d) Liberals: 484(.227) =109.87
(e) Moderates: 610(.227)=138.47
Conservatives: 591(.227)=134.16
(f) 382(484)/1685=109.73
(g) rest of row is 138.47 and 134.16
(h) (74-109.87)2/109.87 = 11.632
(139-138.47)2/138.47 = .004
(169-134.16)2/134.16 =9.152
(410-374.27)2/374.27 =3.140
(471-471.71)2/471.71 = .001
(422-457.02)2/457.02 = 2.683
sum = 26.881
(i) large values of the test statisitc will be evidence against the null hypothesis (indicate larger discrepancies between the observed and the expected)
(j) df = (3-1)(2-1) = 2
26.881 is off the chart so p-value is < .0005
(k) It would be highly unusual to observe these sample data by chance alone if there was independence between agreement and political leaning. We have very strong evidence that political leaning is related to people’s agree with the statement "Most men are better suited emotionally for politics than are
most women."

Activity 26-2: Government Spending

(a) Ho: There is no association between opinionon spending and political leaning
Ha: There is an association between opinionon spending and political leaning
(b) 133(376)/1235 = 40.49
590(435)/1235 = 207.81
512(424)/1235 = 175.78
(c) (40.49-50)2/50 = 2.232
(214-207.81)2/207.81 = .184
(176-175.78)2/175.78 = .000
 sum=6.724
(d) with df=(3-1)(3-1)=4, find .1 < p-value < .2
(e) There is no evidence of an association between opinion and spending on space program.

Activity 26-3: Newspaper Reading (cont.)

(a) MTB > chisq c1 c2
Chi-Square Test: men, women
Expected counts are printed below observed counts
           men    women    Total
    1      375      430      805
        349.12   455.88
    2      182      238      420
        182.15   237.85
    3      121      173      294
        127.50   166.50
    4      133      218      351
        152.23   198.77
Total      811     1059     1870

Chi-Sq =  1.918 +  1.469 +
          0.000 +  0.000 +
          0.332 +  0.254 +
          2.428 +  1.859 = 8.261
DF = 3, P-Value = 0.041

Ho: There is no association between gender and newspaper reading
Ha: There is an association between gender and newspaper reading
test statistic: 8.261, p-value = .041
We would reject the null hypothesis at the .05 level and conclude there is a relationship between gender and newspaper reading. (We would not reject at the .01 level.)
(b) The 2.428 is the largest. This corresponds to the male, reading less than once per week cell. There are fewer males (133) reading less than once per week
than we would’ve expected (152.23) if there was no relationship.
(c) Next three: 1.918, 1.859, 1.469
There are more men than we would have expected reading every day, there are fewer women then we would expected reading every day and more women
than expected reading less than once per week. Seems men read more than women.

Activity 26-4: Suitability for Politics (cont.)

(a)Ho: There is no association between gender and suitability opinion
Ha: There is an association between gender and suitability opinion
MTB > chisq c4 c5
Chi-Square Test: men, women
Expected counts are printed below observed counts
           men    women    Total
    1      169      236      405
        170.16   234.84
    2      565      777     1342
        563.84   778.16
Total      734     1013     1747

Chi-Sq =  0.008 +  0.006 +
          0.002 +  0.002 = 0.018
DF = 1, P-Value = 0.894

 With our large p-value, there is no evidence of an association between gender and their opinion on whether men are more suitable for politics than women.

(b) Let q1=proportion of men who agree with the statement and q2=proportion of women who agree with the statement
Ho: q1=q2 (men and women agree in equal proportions)
Ha: q1¹q2(the proportions of men and women who agree differs)
MTB> PTwo 405 169 1342 565;
SUBC>   Pooled.
Test and CI for Two Proportions
Sample      X      N  Sample p
1         169    405  0.417284
2         565   1342  0.421013
Estimate for p(1) - p(2):  -0.00372946
95% CI for p(1) - p(2):  (-0.0585395, 0.0510806)
Test for p(1) - p(2) = 0 (vs not = 0):  Z = -0.13  P-Value = 0.894
Again, with this large p-value, we would fail to reject Ho: q1=q2, indicating that men and women agree in the same proportion.
(c) The p-values are equal, and the conclusion is the same.
(d) z=-.13, z2=.0169 which is close to .018.
 

Activity 26-5: Government Spending (cont.)

Note: The answers to (a) and (b) below comprise the answer to (a) in the Calculator version. The answers to (c)-(f) below are the answers to (b)-(e) in the Calculator version.
Answers will vary. These are sample answers.
(a) Sample output:
liberal moderate conservative
too little 41 41 51
just right 180 217 193
too much 155 177 180
(b) chi-square value for above table: 2.386
(c) This is smaller than 6.724
(d)

(e) There were 20 values in this graph larger than 6.725. This is 20% of the repetitions.
(f) Earlier found p-value to be between .1 and .2 so we are close.  They should match since the p-value tells us how often we expect to get a chi-square value at least this extreme when the null hypothesis is true.