An instructor is teaching two sections of the same basic statistics course. The instructor is giving the same exams, homework as
1 answer:
Answer:
Two-sample t test.
Because we have two different sections and we want to see if exists difference between the two averages. Both sections are independent so it can be considered as different samples.
Step-by-step explanation:
Two-sample t test.
Because we have two different sections and we want to see if exists difference between the two averages. Both sections are independent so it can be considered as different samples.
If we don't know the population deviations, so for this case is better apply a two sample t test to compare means, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
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Line A:
y - y₁ = m(x - x₁)
y - 1 = 1(x - (-5))
y - 1 = 1(x + 5)
y - 1 = 1(x) + 1(5)
y - 1 = x + 5
+ 1 + 1
y = x + 6
Line B:
y - y₁ = m(x - x₁)
y - (-1) = -3(x - 4)
y + 1 = -3(x) + 3(4)
y + 1 = -3x + 12
- 1 - 1
y = -3x + 11
y = x + 6
y = -3x + 11
x + 6 = -3x + 11
+ 3x + 3x
4x + 6 = 11
- 6 - 6
4x = 5
4 4
x = 1¹/₄
y = x + 6
y = 1¹/₄ + 6
y = 7¹/₄
(x, y) = (1¹/₄, 7¹/₄)
The answer is B.
y - y₁ = m(x - x₁)
y - 1 = 1(x - (-5))
y - 1 = 1(x + 5)
y - 1 = 1(x) + 1(5)
y - 1 = x + 5
+ 1 + 1
y = x + 6
Line B:
y - y₁ = m(x - x₁)
y - (-1) = -3(x - 4)
y + 1 = -3(x) + 3(4)
y + 1 = -3x + 12
- 1 - 1
y = -3x + 11
y = x + 6
y = -3x + 11
x + 6 = -3x + 11
+ 3x + 3x
4x + 6 = 11
- 6 - 6
4x = 5
4 4
x = 1¹/₄
y = x + 6
y = 1¹/₄ + 6
y = 7¹/₄
(x, y) = (1¹/₄, 7¹/₄)
The answer is B.