1. The data below shows the weight and chest size of infants at birth. Weight, (kg) Chest size (cm) (Y) 3.20 4.04 Σ= 3.41 Weight (X) 3.20 4.92 3.17 2.47 2.41 4.04 5.15 3.23 3.41 3.28 4.92 3.17 2.47 2.41 5.15 3.23 3.28 Chest Size (Y) Σ= 26.7 28.5 32.1 26.8 24.9 26.8 32.1 36.8 27.7 28.9 Ha: (in words) H₂: (in symbols) p #0 28.5 I. Statement of Hypothesis: Ho: (in words) Ho: (in symbols) p = 0 III. Test Statistic: 32.1 Two-tailed test Σ= IV. Computation: XY Pearson r nΣxy-ExΣy √√[n(x²)-([x³]In(Σy²³)-([y)²] V. Decision and Conclusion: r pearson R² = 26.7 32.1 26.8 24.9 26.8 spearman 36.8 27.7 28.9 Σ= II. Level of Significance and Critical Value a = 0.05 df = n-2 = CV = Decision Rule: Reject Ho if t-computed is... = X² Compute the Pearson correlation coefficient. Test the hypothesis that there exist no significant relationships between the weight and chest size of the infants. Use 0.05 level of significance. Determine the coefficient of multiple determination and interpret the result. Compute the Spearman correlation coefficient. Σ= Y² Spearman's Rank 6Σd²) n(n²-1) 7₂ = 1- Rank (X) Formula: t= Rank (Y) Statistical Test: t-test for the significance of r r√√n-2 1-r² d Σ= d²

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1. The data below shows the weight and chest size of infants at birth.
Weight, (kg)
Chest size (cm)
(Y)
3.20
4.04
ΣΕ
3.41
Weight (X)
3.20
4.92
3.17
2.47
2.41
4.04
5.15
3.23
3.41
3.28
4.92
3.17
2.47
2.41
5.15
3.23
3.28
Chest Size (Y)
r=
26.7
28.5
32.1
26.8
24.9
26.8
32.1
36.8
27.7
28.9
28.5
III. Test Statistic:
32.1
I. Statement of Hypothesis:
Ho: (in words)
Ho: (in symbols) p = 0
Ha: (in words)
Ha: (in symbols) p #0
Two-tailed test
ΣΕ
IV. Computation:
XY
V. Decision and Conclusion:
Pearson r
nΣxy-ExΣy
√[n([x²)-([x)³][n(Σy²)−(Σy)³]
pearson
R² =
r
26.7
spearman
32.1
26.8
24.9
26.8
36.8
27.7
28.9
II. Level of Significance and Critical Value
a = 0.05
df = n-2 =
CV =
Decision Rule: Reject Ho if t-computed is...
Σ=
X²
Compute the Pearson correlation coefficient.
Test the hypothesis that there exist no significant
relationships between the weight and chest size of
the infants. Use 0.05 level of significance.
Determine the coefficient of multiple determination
and interpret the result.
Compute the Spearman correlation coefficient.
Σ=
Y²
Spearman's Rank
6(Σd²)
n(n² - 1)
7 = 1-.
Rank (X)
Formula: t=
Rank (Y)
Statistical Test: t-test for the significance of r
r√√n-2
d
Σ=
d²
Transcribed Image Text:1. The data below shows the weight and chest size of infants at birth. Weight, (kg) Chest size (cm) (Y) 3.20 4.04 ΣΕ 3.41 Weight (X) 3.20 4.92 3.17 2.47 2.41 4.04 5.15 3.23 3.41 3.28 4.92 3.17 2.47 2.41 5.15 3.23 3.28 Chest Size (Y) r= 26.7 28.5 32.1 26.8 24.9 26.8 32.1 36.8 27.7 28.9 28.5 III. Test Statistic: 32.1 I. Statement of Hypothesis: Ho: (in words) Ho: (in symbols) p = 0 Ha: (in words) Ha: (in symbols) p #0 Two-tailed test ΣΕ IV. Computation: XY V. Decision and Conclusion: Pearson r nΣxy-ExΣy √[n([x²)-([x)³][n(Σy²)−(Σy)³] pearson R² = r 26.7 spearman 32.1 26.8 24.9 26.8 36.8 27.7 28.9 II. Level of Significance and Critical Value a = 0.05 df = n-2 = CV = Decision Rule: Reject Ho if t-computed is... Σ= X² Compute the Pearson correlation coefficient. Test the hypothesis that there exist no significant relationships between the weight and chest size of the infants. Use 0.05 level of significance. Determine the coefficient of multiple determination and interpret the result. Compute the Spearman correlation coefficient. Σ= Y² Spearman's Rank 6(Σd²) n(n² - 1) 7 = 1-. Rank (X) Formula: t= Rank (Y) Statistical Test: t-test for the significance of r r√√n-2 d Σ= d²
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