Question 1. Highest Mean Salary
The position with the highest mean salary was the QB position, which had a mean salary of 4.041e+6.
Figure 1
M ean Salaries
Question 2. Sampling
The sample selected was 300, as it represented 15% of the total population of 2000. The mean of the population was 1.789e+6 and a standard deviation of 2.228e+6.
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Table 1 Descriptive Statistics |
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Total Salary |
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Valid |
300 |
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Missing |
0 |
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Mean |
1.789e +6 |
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Std. Deviation |
2.228e +6 |
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Minimum |
310000.000 |
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Maximum |
1.401e +7 |
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The sample size is adequate as it represented a significant percentage of the population. Smaller sample size would have increased the error in the result. The sampling method was appropriate as it was truly random. The function “Rand ()” identified random data from the sample and did not introduce any bias. The mean salary for players was determined as $1,789,000. The implication is that players, agents, teams should consider this as the average salary. The amount will be critical when negotiating about player earnings.
Task 2 Calculating Confidence Intervals
1. Current Donation Level Amount
The mean for the annual contribution was 6,212.057, and it had a standard deviation of 29,657.971. The mean for lifetime contributions was 82,010.188, and the standard deviation was 434,046.645. The values were as shown in table 2.
Table 2 Descriptive Statistics |
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Contribute Annually |
Contribute Lifetime |
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Valid |
597 |
584 |
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Missing |
47 |
60 |
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Mean |
6212.047 |
82010.188 |
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Std. Deviation |
29657.971 |
434046.645 |
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Minimum |
100.000 |
50.000 |
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Maximum |
500000.000 |
1.000e +7 |
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The formula for calculating the confidence interval is:
CI = x̅ ± z * ( σ / )
Where x̅ is the sample mean , z appropriate z-value from the normal distribution table, σ is the standard deviation, and n is the sample size. The z-value for a 95% confidence interval is 1.96.
For the annual giving amount,
x̅ = 6,212.057, z = 1.96, σ = 29,657.971, and n = 597.
CI = 6,212.057 ± 1.96 * ( 29,657.971 / )
CI = 6,212.057 ± 2379.08
Upper = 8591.137, lower = 3832.977
Standard error (SE) = Upper limit – Lower Limit)/z value
Z value is (3.92 for 95% CI)
SE = (8591.137 - 3832.977)/3.92 = 1213.82
SE = 1213.82
For lifetime giving amount x̅ = 82,010.188 , z = 1.96, σ = 434,046.645, and n = 584.
CI = 82,010.188 ± 1.96 * ( 434046.645/ )
CI = 82,010.188 ± 35203.51
Upper = 117,213.698, Lower = 46,806.678
SE = ( 82,010.188 - 46,806.678)/3.92
SE = 8980.48
2. Additional Comments
The mean annual contribution was calculated as 6,212.057, and the highest amount could be 8591.137 with the lowest 3832.977. The executive management should consider the amount when considering the budget for future years. The mean lifetime contributions were 82010.188, and the highest amount could be 117,213.698 with the lowest amount being 46,806.678. The lifetime contributions were significantly high, and the management should consider this as loyalty points for players.
Distribution Method
Table 4. Annual Contributions Estimated Parameters |
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95% CI |
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Parameter |
Estimate |
SE |
Lower |
Upper |
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μ |
6212.130 |
1210.791 |
3839.022 |
8585.237 |
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σ² |
8.781e +8 |
5.074e +7 |
7.787e +8 |
9.776e +8 |
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Note. Standard errors and confidence intervals were calculated using the delta method. |
Table 5. Lifetime Contributions Estimated Parameters |
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95% CI |
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Parameter |
Estimate |
SE |
Lower |
Upper |
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μ |
82011.638 |
1482.910 |
79105.187 |
84918.089 |
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σ² |
1.881e +11 |
1.286e +9 |
1.856e +11 |
1.906e +11 |
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Note. Standard errors and confidence intervals were calculated using the delta method. |
Comparison of Values
Values from calculation were compared to that in the distribution method. The comparison was shown in table 6.
Table 6.
Calculation vs. Distribution Method
Calculation | Distribution Method | |
Annual Contribution | ||
SE | 1213.82 | 1210.791 |
Lower | 3832.977 | 3839.022 |
Upper | 8591.137 | 8585.237 |
Lifetime Contribution | ||
SE | 8980.48 | 1 , 482.910 |
Lower | 46,806.678 | 79 , 105.187 |
Upper | 117,213.698 | 84 , 918.089 |
The confidence interval and standard error values were similar in the annual contributions but different in lifetime contributions. The difference in the values could have been due to the difference in the formula applied.