Question

Using the MLB attendance data, calculate 90% Confidence leveIs for each team mean attendance. What sampling distribution did you use? Why? Can you conclude that one team or another has larger crowds given these Confidence leveIs?

Brewers White Sox             Indians   Twins

45,341    38,088           16,789    48,711

22,603    26,337           18,082    24,439

23,649    24,141           14,887    27,539

41,758    33,278           13,843    26,047

41,282    27,653           14,066    24,552

28,019    23,139           20,484    30,131

22,331    25,390           25,065    15,869

17,386    25,459           25,402    27,783

41,522    26,342           25,721    35,269

41,209    30,193           27,250    27,024

31,985    38,208           41,675    16,076

20,191    38,513           41,977    20,849

42,058    32,315           41,612    21,496

41,704    33,147           37,570    18,520

39,483    36,399           39,250    24,880

42,126    38,586           23,178    31,664

43,716    38,210           24,784    33,663

37,518    36,629           27,303    38,334

33,404    35,391           25,949    28,314

36,062    35,309           38,325    30,924

41,008    31,739           30,410    41,037

43,087    37,639           34,372    36,353

42,398    30,581           32,624    31,755

35,689    38,874           31,599    42,373

37,690    36,745           29,657    29,881

42,559    37,030           21,460    24,986

42,754    35,161           21,321    15,736

36,381    37,281           29,822    25,207

32,540    33,640           32,439    36,338

30,247    30,364           39,339    36,066

41,156    32,426           29,614    36,737

43,121    31,543           34,286    30,939

43,180    36,791           37,292    38,070

42,554    32,724           41,203    36,551

41,790    30,122           37,102    35,794

30,852    30,569           21,811    33,868

42,250    29,042           41,365    33,217

43,172    28,462           38,254    23,628

35,760    30,567           30,628    22,890

38,535    34,590           14,036    25,144

40,186    37,744           17,632    39,742

41,631    36,571           19,315    31,035

35,238    32,703           22,325    25,868

35,151    30,895           34,849    27,903

36,328    30,829           35,153    26,714

41,721    32,688           33,429    27,977

44,064    33,212           18,710    37,117

28,789    36,616           17,371    31,624

30,713    33,433           24,278    24,240

31,862    36,132           17,737    27,000

20,446    35,327           18,494    30,959

23,229    32,007           18,614    31,038

15,602    39,046           22,921    34,951

40,190    39,043           34,557    40,088

40,361    39,194           36,726    27,807

37,761    34,522           13,389    21,979

17,751    36,702           14,163    24,367

19,398    34,468           16,284    23,663

24,658    30,953           23,325    31,458

44,759    34,609           17,678    32,176

44,427    30,488           28,609    25,037

39,119    35,327           34,230    26,781

41,139    34,122           35,262    31,434

27,559    29,051           32,524    25,781

32,758    34,538           38,645    27,090

17,704    38,103           30,038    23,771

34,190    34,104           41,131    29,042

39,339    31,747           37,718    21,738

31,226    32,250           35,230    26,454

25,854    30,126           32,111    24,105

28,988    31,939           30,112    16,128

42,944    23,537           28,831    13,977

40,710    33,581           41,103    22,282

31,150    36,485           32,511    35,230

40,908    29,010           36,016    21,771

32,329    31,607           40,663    14,197

32,411    32,091           40,250    18,226

34,918    34,477                  17,842

38,135    30,281                  17,982

40,946    33,066                  31,737

42,415    33,154                  29,382

Answer 1

We know that, by the central limit theorem, the sampling distribution of the mean of any independent, random variable will be normal or nearly normal, if the sample size is large enough.

here we have n=81 which is large enough to state that, we can use normal distribution as a sampling distribution of each team mean attendance to calculate 90% Confidence leveIs.

#90% CI.
#This means alpha = .10 We can get z(alpha/2) = z(0.05)
> qnorm(.95)
[1] 1.644854
> mean(br)
[1] 35420.7
> sd(br)
[1] 7899.475

team Brewers
> mean(br)-qnorm(.95)*(sd(br)/length(br))
[1] 35260.29
> mean(br)+qnorm(.95)*(sd(br)/length(br))
[1] 35581.12

team Indians

> #lower bound
> mean(ind)-qnorm(.95)*(sd(ind)/length(ind))
[1] 28694.22
> #upper bound
> mean(ind)+qnorm(.95)*(sd(ind)/length(ind))
[1] 29066.3

team twins

> #lower bound
> mean(tw)-qnorm(.95)*(sd(tw)/length(tw))
[1] 28082.89
> #upper bound
> mean(tw)+qnorm(.95)*(sd(tw)/length(tw))
[1] 28370.12

team white socks

> #lower bound
> mean(wh)-qnorm(.95)*(sd(wh)/length(wh))
[1] 33061.95
> #upper bound
> mean(wh)+qnorm(.95)*(sd(wh)/length(wh))
[1] 33219.41

As for team India , the margin of error i.e z(alpha/2) * sigma/sqrt(n) this quantity is highest among all 4 teams, team Indian has larger crowd.