Question

In: Statistics and Probability

Using the data below, perform descriptive statistics analyses (mean, standard deviation, variance) for scores from the...

Using the data below, perform descriptive statistics analyses (mean, standard deviation, variance) for scores from the two class periods. Then,determine if there is a significant difference between the two sets of exam scores by performing an independent t test. Lastly, write a paragraph to convey the findings, using APA format.

The number would not align as I like but Third Period Scores are as follow: 84, 92,,98,76,80,90,70,94,100,94,98,88,90,92,88,76,80,92 and 88

Sixth Period Scores: 80,90,82,78,72,80,70,86,80,88,70,86,68,76,90,82,70,84, and 84

Please write legible to understand response on the answer.

Thanking you in advance

Expert Solution

For Third Period Scores:

Mean = 84 + 92 + 98 + 76 + 80 + 90 + 70 + 94 + 100 + 94 + 98 + 88 + 90 + 92 + 88 + 76 + 80 + 92 + 88/19 = 87.89

Standard deviation = √(84 - 87.89)² + (92 - 87.89)² + (98 - 87.89)² + (76 - 87.89)² + (80 - 87.89)² + (90 - 87.89)² + (70 - 87.89)² + (94 - 87.89)² + (100 - 87.89)² + (94 - 87.89)² + (98 - 87.89)² + (88 - 87.89)² + (90 - 87.89)² + (92 - 87.89)² + (88 - 87.89)² + (76 - 87.89)² + (80 - 87.89)² + (92 - 87.89)² + (88 - 87.89)²/19 - 1 = 8.26

Variance = Standard deviation² = 8.26² = 68.21

For Sixth Period Scores:

Mean = 80 + 90 + 82 + 78 + 72 + 80 + 70 + 86 + 80 + 88 + 70 + 86 + 68 + 76 + 90 + 82 + 70 + 84 + 84/19 = 79.79

Standard deviation = √(80 - 79.79)² + (90 - 79.79)² + (82 - 79.79)²+ (78 - 79.79)² + (72 - 79.79)² + (80 - 79.79)² + (70 - 79.79)² + (86 - 79.79)² + (80 - 79.79)² + (88 - 79.79)² + (70 - 79.79)² + (86 - 79.79)² + (68 - 79.79)² + (76 - 79.79)² + (90 - 79.79)² + (82 - 79.79)² + (70 - 79.79)² + (84 - 79.79)² + (84 - 79.79)²/19 - 1 = 7.08

Variance = Standard deviation² = 7.08² = 50.18

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

s² = 18*68.21 + 18*50.18/19 + 19 - 2 = 59.193

t = (87.89 - 79.79)/√59.193(1/ 19 + 1/19) = 3.247

df = 19 + 19 - 2 = 36

The p-value for df = 36 and t = 3.247 is 0.0025.

Since the p-value (0.0025) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that there is a significant difference between the two sets of exam scores.

orchestra answered 2 years ago

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