1)The margin of error requires the number of observations use, =COUNT(F2:F55), which equals 54

Standard deviation: =STDEV(F2:F55) = 7.413348

Mean: =AVERAGE(F2:F55) = 16.2037

The margin of error is found by =CONFIDENCE.T(1-0.99, 7.413348 , 54), or about

2.695412

The 90% confidence interval estimate for the average number of points the losing team scores is then

[16.2037- 2.695412, 16.2037+ 2.695412] or [13.50829, 18.89912]

2)Average of winning points: =AVERAGE(D2:D55)= 30.11111

Null: the winning teams, on average, scores points is equal to the upper bound of the confidence interval from #1 for the losing teams

H₀: ?_win = ?_lose

Alt: the average of winning points and the upper bound of the confidence interval from #1 for the losing teams are different

H_A: ?_win ?_lose

3) z= 8.33

4) p= 0.000

5) Since the p value < 0.01 so we should reject the null hypothesis and conclude that winning team on average score more than loser team 's upper limit as (the sample average for winning team is much more than the upper limit of loser team)

Please answer #6

6). What does the findings from #1 and #5 tell you? Explain them in words a person who has not taken a statistics course could understand.