Answer:
Mid-point:
Equation:
Step-by-step explanation:
To find the mid-point of AB, simply add up their x and y coordinates and divide by 2 respectively to find their middle point.
To find the perpendicular slope that passes through the mid-point, we need to know the slope between AB first.
Slope of AB: =
=
=
Multiplying slopes that are perpendicular with each other always results in -1.
By the point slope form:
Plug in the coordinates of the mid-point:
Equation:
It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.
<h3>What is a t test?</h3>
A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.
<h3>What are the conditions for using this kind of confidence interval? </h3>The conditions to use the t test are:
The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:
df = [(s₁²/n₁) + (s₂²/n²)/[((s₁²/n₁)²/((n₁-1)) + (s₂²/n₂)²/((n₂-1))]
= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]
= 24.965
Hence,
df ≈ 24 (if approximated to the floor)
<h3>What are the sample statistics for this test?</h3>Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.
Here, H₀ is given as;
t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]
= - 0.8/√0.6337
t = - 1.005
<h3>What is the 95% confidence interval for the difference between the number of classes missed by each group of students?</h3>
The 95% confidence interval is computed using the following formula:
= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)
= - 0.8 ± 2.064 (√0.6337)
= -2.4429, 0.8429
<h3>What is the a 90% confidence interval for the difference between the number of classes missed by each group of students?</h3>To derive the 90% interval, we state:
= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)
= - 0.8 ± 0.685 (√0.6337)
= -2.162, 0.562
<h3>Based on the two confidence intervals computed in parts d and e, what is the conclusion about the differences between the means of the two groups?</h3>
From the intervals computed, we must fail to reject H₀
H₀ : μ₁ = μ₂
It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.
Learn more about t-test at;
brainly.com/question/6589776
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