In: Statistics and Probability
The baseball commissioner believes that the average attendance is less than 2,300,000 per team. You decide to conduct a test of hypothesis to determine whether the mean attendance (Attendance Column) was more than 2,300,000 per team. Use the 5% level of significance. Note: We do not know the populations standard deviation.
| Attendance | 
| 2.16 | 
| 2.97 | 
| 2.68 | 
| 2.28 | 
| 3.05 | 
| 1.62 | 
| 3.37 | 
| 2.30 | 
| 4.27 | 
| 1.92 | 
| 2.67 | 
| 1.39 | 
| 2.35 | 
| 2.36 | 
| 2.32 | 
| 2.75 | 
| 3.25 | 
| 2.06 | 
| 2.38 | 
| 1.37 | 
| 3.02 | 
| 3.86 | 
| 2.87 | 
| 3.85 | 
| 3.11 | 
| 1.75 | 
| 2.79 | 
| 3.22 | 
| 3.55 | 
| 1.96 | 
Select one:
a. p = .700%. Reject the null. The mean attendance is more than 2.3 million.
b. p = 99.30%. Reject the null. The mean attendance is more than 2.3 million.
c. p = 99.30%. Do not reject the null. The mean attendance is less than 2.3 million.
d. p = .700%. Do not reject the null. The mean attendance is less than 2.3 million.


the test statistic we used here is

| S NO | Attendance(x) | ![]()  | 
![]()  | 
| 1 | 2.16 | -0.49 | 0.2401 | 
| 2 | 2.97 | 0.32 | 0.1024 | 
| 3 | 2.68 | 0.03 | 0.0009 | 
| 4 | 2.28 | -0.37 | 0.1369 | 
| 5 | 3.05 | 0.4 | 0.16 | 
| 6 | 1.62 | -1.03 | 1.0609 | 
| 7 | 3.37 | 0.72 | 0.5184 | 
| 8 | 2.3 | -0.35 | 0.1225 | 
| 9 | 4.27 | 1.62 | 2.6244 | 
| 10 | 1.92 | -0.73 | 0.5329 | 
| 11 | 2.67 | 0.02 | 0.0004 | 
| 12 | 1.39 | -1.26 | 1.5876 | 
| 13 | 2.35 | -0.3 | 0.09 | 
| 14 | 2.36 | -0.29 | 0.0841 | 
| 15 | 2.32 | -0.33 | 0.1089 | 
| 16 | 2.75 | 0.1 | 0.01 | 
| 17 | 3.25 | 0.6 | 0.36 | 
| 18 | 2.06 | -0.59 | 0.3481 | 
| 19 | 2.38 | -0.27 | 0.0729 | 
| 20 | 1.37 | -1.28 | 1.6384 | 
| 21 | 3.02 | 0.37 | 0.1369 | 
| 22 | 3.86 | 1.21 | 1.4641 | 
| 23 | 2.87 | 0.22 | 0.0484 | 
| 24 | 3.85 | 1.2 | 1.44 | 
| 25 | 3.11 | 0.46 | 0.2116 | 
| 26 | 1.75 | -0.9 | 0.81 | 
| 27 | 2.79 | 0.14 | 0.0196 | 
| 28 | 3.22 | 0.57 | 0.3249 | 
| 29 | 3.55 | 0.9 | 0.81 | 
| 30 | 1.96 | -0.69 | 0.4761 | 
| Total | 79.5 | 15.5414 | |
| Mean | 2.65 | ||
| S square | 0.5359 | ||
| s | 0.7321 | 
mean



Variance




now


the Pvalue of Z = 2.6185 is 0.004416
since the p value is less than alpha= 0.05 therefore we reject the null hypothesis and claim that mean is greater than 2.3 million