Math Problem Statement

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below.

Observation

1

2

3

4

5

6

7

8   Upper X Subscript iXi

44.644.6

43.543.5

44.644.6

42.642.6

44.444.4

46.146.1

49.849.8

50.250.2

Upper Y Subscript iYi

48.148.1

45.145.1

47.947.9

47.347.3

45.845.8

46.246.2

51.251.2

52.352.3

Question content area bottom

Part 1

​(a) Determine

d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi

for each pair of data.

Observation

1

2

3

4

5

6

7

8

d Subscript i

negative 3.5−3.5

negative 1.6−1.6

negative 3.3−3.3

negative 4.7−4.7

negative 1.4−1.4

negative 0.1−0.1

negative 1.4−1.4

negative 2.1−2.1

​(Type integers or​ decimals.)

Part 2

​(b) Compute

d overbard

and

s Subscript dsd.

d overbardequals=negative 2.263−2.263

​(Round to three decimal places as​ needed.)

Part 3

s Subscript dsdequals=enter your response here

​(Round to three decimal places as​ needed.)

Solution

Let's compute the required values step by step.


Part (a): Calculate di=XiYid_i = X_i - Y_i

The did_i values for each observation are already provided: d1=3.5,d2=1.6,d3=3.3,d4=4.7,d5=1.4,d6=0.1,d7=1.4,d8=2.1d_1 = -3.5, \, d_2 = -1.6, \, d_3 = -3.3, \, d_4 = -4.7, \, d_5 = -1.4, \, d_6 = -0.1, \, d_7 = -1.4, \, d_8 = -2.1


Part (b): Compute dˉ\bar{d} (mean of did_i)

To calculate dˉ\bar{d}, use the formula: dˉ=din\bar{d} = \frac{\sum d_i}{n} where n=8n = 8 (number of observations).

di=(3.5)+(1.6)+(3.3)+(4.7)+(1.4)+(0.1)+(1.4)+(2.1)\sum d_i = (-3.5) + (-1.6) + (-3.3) + (-4.7) + (-1.4) + (-0.1) + (-1.4) + (-2.1)

di=18.1\sum d_i = -18.1

dˉ=18.18=2.263\bar{d} = \frac{-18.1}{8} = -2.263

Thus, dˉ=2.263\bar{d} = -2.263 (rounded to three decimal places).


Part (c): Compute sds_d (standard deviation of did_i)

To calculate sds_d, use the formula: sd=(didˉ)2n1s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}

Step 1: Compute (didˉ)2(d_i - \bar{d})^2 for each did_i

d1dˉ=3.5(2.263)=1.237,(1.237)2=1.530d_1 - \bar{d} = -3.5 - (-2.263) = -1.237, \quad (-1.237)^2 = 1.530 d2dˉ=1.6(2.263)=0.663,(0.663)2=0.440d_2 - \bar{d} = -1.6 - (-2.263) = 0.663, \quad (0.663)^2 = 0.440 d3dˉ=3.3(2.263)=1.037,(1.037)2=1.076d_3 - \bar{d} = -3.3 - (-2.263) = -1.037, \quad (-1.037)^2 = 1.076 d4dˉ=4.7(2.263)=2.437,(2.437)2=5.939d_4 - \bar{d} = -4.7 - (-2.263) = -2.437, \quad (-2.437)^2 = 5.939 d5dˉ=1.4(2.263)=0.863,(0.863)2=0.745d_5 - \bar{d} = -1.4 - (-2.263) = 0.863, \quad (0.863)^2 = 0.745 d6dˉ=0.1(2.263)=2.163,(2.163)2=4.678d_6 - \bar{d} = -0.1 - (-2.263) = 2.163, \quad (2.163)^2 = 4.678 d7dˉ=1.4(2.263)=0.863,(0.863)2=0.745d_7 - \bar{d} = -1.4 - (-2.263) = 0.863, \quad (0.863)^2 = 0.745 d8dˉ=2.1(2.263)=0.163,(0.163)2=0.027d_8 - \bar{d} = -2.1 - (-2.263) = 0.163, \quad (0.163)^2 = 0.027

Step 2: Compute (didˉ)2\sum (d_i - \bar{d})^2

(didˉ)2=1.530+0.440+1.076+5.939+0.745+4.678+0.745+0.027=15.180\sum (d_i - \bar{d})^2 = 1.530 + 0.440 + 1.076 + 5.939 + 0.745 + 4.678 + 0.745 + 0.027 = 15.180

Step 3: Compute sds_d

sd=15.18081=15.1807=2.16857=1.472s_d = \sqrt{\frac{15.180}{8 - 1}} = \sqrt{\frac{15.180}{7}} = \sqrt{2.16857} = 1.472

Thus, sd=1.472s_d = 1.472 (rounded to three decimal places).


Final Results:

  1. dˉ=2.263\bar{d} = -2.263
  2. sd=1.472s_d = 1.472

Would you like further details or clarifications on these computations?


Expanding Questions:

  1. How would the results change if one additional observation was included?
  2. How does the assumption of normality affect the interpretation of these calculations?
  3. Can you calculate a confidence interval for dˉ\bar{d} given sds_d and the sample size?
  4. How would you test whether dˉ\bar{d} is significantly different from zero?
  5. What is the practical interpretation of dˉ\bar{d} and sds_d in a real-world context?

Tip:

Always check for outliers in your data, as they can heavily influence measures like the mean and standard deviation.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Analysis
Normal Distribution

Formulas

d̄ = (Σdi) / n
sd = sqrt((Σ(di - d̄)^2) / (n - 1))

Theorems

Central Limit Theorem (underlying assumption of normality)

Suitable Grade Level

Grades 10-12, College Level