Question

(4+4+2 = 10 points) Construct a mosaic plot for “League” versus “Division” with the corresponding contingency table. What is the probability that a randomly selected player played in League “A” given that s/he also played in Division “E”?

From ISLR ("Hitters") data package

Answer 1

Rcode:

library(ISLR)

library(gmodels)

table = xtabs(~League+Division, data = Hitters)
mosaicplot(table, shade = TRUE)
CrossTable(Hitters$League, Hitters$Division,chisq = TRUE, expected = TRUE, sresid = TRUE, format = "SPSS")

Output:

League Residuals: Standardized Division ------------ >2020-24 table

cell Contents --- ------- ----- --------- Count Expected values | chi-square contribution ROW Percent column Percent Total Percent Std Residual |

------------------------- Total observations in Table: 322 secher 107 obs. of 4 variables values Europe_hambgnum (1:12] 5.14 4.93 4.07 4.64 5 Female num (1:87 24018 13405 18218 1734 Files Plots Packages Help Viewer Zoom Export- table - - A N - - | Hitters Division Hitters $League | EL W Row Total | -------- ----------- ----------- 85 90 175 85.326 L 89.674 L 0.001 0.001 48. 571% 51.429% 54. 348% 54.140% 54.545% 26. 398% 27.950% -0.035 0.034 1 - - ------- ---- ---- ------- -- - 72 75 147 71.674 75.326 0.001 L 0.001 | 48.980% 51. 020% 45.652% 45.860% 45.455% 22.360% 23. 292% 0.039 -0.038 ---- - ---------- -------- --- ---- ------- -- -- column Total | 157 165 48.758% 51. 242% -- ---- ---- --- Division Standardized Residuals: League tivate Window

Statistics for All Table Factors Pearson's chi-squared test chi^2 = 0.005327157 d. f. = 1 p = 0.9418162 Pearson's chi-squared test with Yates' continuity correction chi^2 = 0 d. f. = 1 p = 1 Minimum expected frequency: 71.67391

Intrepretation

From Mosaic plot we could see there are no signiifcant differences between league and division

also from chisq test statisitic

Chi^2 = 0.005327157 and p = 0.9418162

p>0.05

Fail to reject Ho

There is no relationship between league and division