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Sagot :
To solve this question, we'll go through the following steps:
1. Calculate the differences between the scores with and without the coach for each player.
Given data:
- "With" scores: [tex]\([17, 20, 19, 18, 19, 19, 20, 20, 16, 17]\)[/tex]
- "Without" scores: [tex]\([16.5, 18, 19, 17.5, 20, 17.5, 19, 19.5, 14, 15.5]\)[/tex]
The differences (With - Without) are:
[tex]$ [0.5, 2, 0, 0.5, -1, 1.5, 1, 0.5, 2, 1.5] $[/tex]
2. Calculate the mean of the differences.
The mean difference ([tex]\(\bar{x}_{\text{diff}}\)[/tex]) is calculated by taking the sum of all differences and dividing by the number of players (10 in this case).
Sum of differences:
[tex]$ 0.5 + 2 + 0 + 0.5 - 1 + 1.5 + 1 + 0.5 + 2 + 1.5 = 8.5 $[/tex]
Mean difference ([tex]\(\bar{x}_{\text{diff}}\)[/tex]):
[tex]$ \bar{x}_{\text{diff}} = \frac{8.5}{10} = 0.85 $[/tex]
So, the mean difference rounded to 2 decimal places is:
[tex]$ \bar{x}_{\text{diff}} = 0.85 $[/tex]
3. Calculate the standard deviation of the differences.
The standard deviation of the differences ([tex]\(S_{\text{diff}}\)[/tex]) measures the amount of variation or dispersion of the differences from the mean.
Sample standard deviation formula (since we are working with a sample, not a population):
[tex]$ S_{\text{diff}} = \sqrt{\frac{\sum (x_i - \bar{x}_{\text{diff}})^2}{n-1}} $[/tex]
Here, [tex]\( n = 10 \)[/tex], so:
[tex]$ S_{\text{diff}} = \sqrt{\frac{\sum (x_i - 0.85)^2}{10-1}} $[/tex]
Calculations for each difference's squared deviation from the mean:
[tex]\[ \begin{align*} (0.5 - 0.85)^2 &= 0.1225 \\ (2 - 0.85)^2 &= 1.3225 \\ (0 - 0.85)^2 &= 0.7225 \\ (0.5 - 0.85)^2 &= 0.1225 \\ (-1 - 0.85)^2 &= 3.4225 \\ (1.5 - 0.85)^2 &= 0.4225 \\ (1 - 0.85)^2 &= 0.0225 \\ (0.5 - 0.85)^2 &= 0.1225 \\ (2 - 0.85)^2 &= 1.3225 \\ (1.5 - 0.85)^2 &= 0.4225 \\ \end{align*} \][/tex]
Sum of these squared deviations:
[tex]$ 0.1225 + 1.3225 + 0.7225 + 0.1225 + 3.4225 + 0.4225 + 0.0225 + 0.1225 + 1.3225 + 0.4225 = 8.0245 $[/tex]
Now, divide by [tex]\( n-1 = 9 \)[/tex]:
[tex]$ \frac{8.0245}{9} = 0.8916 $[/tex]
Finally, take the square root to get the standard deviation:
[tex]$ S_{\text{diff}} = \sqrt{0.8916} \approx 0.944 $[/tex]
So, the standard deviation rounded to 3 decimal places is:
[tex]$ S_{\text{diff}} = 0.944 $[/tex]
4. Count the number of differences.
Given that there are 10 players, the number of differences ([tex]\(n_{\text{diff}}\)[/tex]) is:
[tex]$ n_{\text{diff}} = 10 $[/tex]
To summarize, we found:
[tex]\[ \begin{align*} \bar{x}_{\text{diff}} &= 0.85 \\ S_{\text{diff}} &= 0.944 \\ n_{\text{diff}} &= 10 \end{align*} \][/tex]
Thus, the final answers are:
[tex]\[ \begin{align*} \bar{x}_{\text{diff}} &= 0.85 \\ S_{\text{diff}} &= 0.944 \\ n_{\text{diff}} &= 10 \end{align*} \][/tex]
1. Calculate the differences between the scores with and without the coach for each player.
Given data:
- "With" scores: [tex]\([17, 20, 19, 18, 19, 19, 20, 20, 16, 17]\)[/tex]
- "Without" scores: [tex]\([16.5, 18, 19, 17.5, 20, 17.5, 19, 19.5, 14, 15.5]\)[/tex]
The differences (With - Without) are:
[tex]$ [0.5, 2, 0, 0.5, -1, 1.5, 1, 0.5, 2, 1.5] $[/tex]
2. Calculate the mean of the differences.
The mean difference ([tex]\(\bar{x}_{\text{diff}}\)[/tex]) is calculated by taking the sum of all differences and dividing by the number of players (10 in this case).
Sum of differences:
[tex]$ 0.5 + 2 + 0 + 0.5 - 1 + 1.5 + 1 + 0.5 + 2 + 1.5 = 8.5 $[/tex]
Mean difference ([tex]\(\bar{x}_{\text{diff}}\)[/tex]):
[tex]$ \bar{x}_{\text{diff}} = \frac{8.5}{10} = 0.85 $[/tex]
So, the mean difference rounded to 2 decimal places is:
[tex]$ \bar{x}_{\text{diff}} = 0.85 $[/tex]
3. Calculate the standard deviation of the differences.
The standard deviation of the differences ([tex]\(S_{\text{diff}}\)[/tex]) measures the amount of variation or dispersion of the differences from the mean.
Sample standard deviation formula (since we are working with a sample, not a population):
[tex]$ S_{\text{diff}} = \sqrt{\frac{\sum (x_i - \bar{x}_{\text{diff}})^2}{n-1}} $[/tex]
Here, [tex]\( n = 10 \)[/tex], so:
[tex]$ S_{\text{diff}} = \sqrt{\frac{\sum (x_i - 0.85)^2}{10-1}} $[/tex]
Calculations for each difference's squared deviation from the mean:
[tex]\[ \begin{align*} (0.5 - 0.85)^2 &= 0.1225 \\ (2 - 0.85)^2 &= 1.3225 \\ (0 - 0.85)^2 &= 0.7225 \\ (0.5 - 0.85)^2 &= 0.1225 \\ (-1 - 0.85)^2 &= 3.4225 \\ (1.5 - 0.85)^2 &= 0.4225 \\ (1 - 0.85)^2 &= 0.0225 \\ (0.5 - 0.85)^2 &= 0.1225 \\ (2 - 0.85)^2 &= 1.3225 \\ (1.5 - 0.85)^2 &= 0.4225 \\ \end{align*} \][/tex]
Sum of these squared deviations:
[tex]$ 0.1225 + 1.3225 + 0.7225 + 0.1225 + 3.4225 + 0.4225 + 0.0225 + 0.1225 + 1.3225 + 0.4225 = 8.0245 $[/tex]
Now, divide by [tex]\( n-1 = 9 \)[/tex]:
[tex]$ \frac{8.0245}{9} = 0.8916 $[/tex]
Finally, take the square root to get the standard deviation:
[tex]$ S_{\text{diff}} = \sqrt{0.8916} \approx 0.944 $[/tex]
So, the standard deviation rounded to 3 decimal places is:
[tex]$ S_{\text{diff}} = 0.944 $[/tex]
4. Count the number of differences.
Given that there are 10 players, the number of differences ([tex]\(n_{\text{diff}}\)[/tex]) is:
[tex]$ n_{\text{diff}} = 10 $[/tex]
To summarize, we found:
[tex]\[ \begin{align*} \bar{x}_{\text{diff}} &= 0.85 \\ S_{\text{diff}} &= 0.944 \\ n_{\text{diff}} &= 10 \end{align*} \][/tex]
Thus, the final answers are:
[tex]\[ \begin{align*} \bar{x}_{\text{diff}} &= 0.85 \\ S_{\text{diff}} &= 0.944 \\ n_{\text{diff}} &= 10 \end{align*} \][/tex]
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