Water and Carbohydrates Continuing on the theme of fruits an

Water and Carbohydrates. Continuing on the theme of fruits and vegetables, here are the number of grams of water and the number of grams of carbohydrates for a random selection of raw foods (100 g each). Is there a linear relationship between the variables?

H_o: ? = 0

H₁: ? not equal 0

Step 2: Find the critical value (from table I) (example: .123)

Critical r value is:

Step 3: Compute the test value using the formula or calculator (round to three decimal places, example .645):

r test value is:

Step 4: Reject the null or do not reject the null (type in either Reject the null or do not reject the null only):

Step 5: Conclusion sentence (type in either isor is not only, to reflect what you found) and state if it is positive or negative.

There        enough evidence to support a       relationship.

Find the equation of the regression line y\' = a + bx and fill in a and b below (round a and b to three decimal places, example: 4.123 or .234)

y\' =

If there are 75 grams of water, how many grams of carbs? (do not round)

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Set Up Hypothesis
Under The Null Hypothesis H0: ? =0
Under The Alternate Hypothesis H1: ?!=0
Test Statistic
Value of ( r ) =-0.9926
Number (n)=6
we use Test Statistic (t) = r / Sqrt(1-r^2/(n-2))
to=-0.9926/(Sqrt( ( 1--0.9926^2 )/(6-2) )
to =-16.35
|to | =16.35
Critical Value
The Value of |t ?| at LOS 0.05% is 2.776
We got |to| =16.35 & | t ? | =2.776
Make Decision
Hence Value of | to | > | t ?| and Here we Reject Ho

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Water	83.93	80.76	87.66	85.20	72.85	84.61	83.81
Carbs	15.25	16.55	11.10	13.01	24.27	14.13	15.11

Correlation X) 83.93 80.76 87.66 85.2 72.85 84.61 83.81 XAz 7044.2449 6522.1776 7684.2756 7259.04 5307.1225 7158.8521 7024.1161 YAz 232.5625 273.9025 123.21 169.2601 589.0329 199.6569 228.3121 X-Y 1279.9325 1336.578 973.026 1108.452 1768.0695 1195.5393 1266.3691 8927.9664 Count 15.25 6.55 11.1 13.01 24.27 14.13 15.11 578.82| 109.42 | (Ctrl) 3288 |1815.937 | |Totals r( x,Y) Co V(X,Y) /S.D (X) S.D (y) r( X,Y)-Sum(XY) / N-Mean of (x) * Mean of (Y)/ Sqrt(X^2/n-( Mean of X)^2 ) Sqrt( Y^2/n-( Cov ( X, Y ) = 1 /7 (8927.9664)-[ 1/7 *578.82 ] [ 1/7 *109.42]--17.117 4.441 S. D ( X ) = Sqrt( 1/7#47999.8288-(1/7*578.82)^2) s.D (Y) = Sqrt( 1/7*1815.937-(1/7*109.42)^2) 3.883 r(x,y) = 7.117 / 4.441*3.883 0.9926 If r-0.9926

Solution

H_o: = 0

H₁: not equal 0

r = -0.9927

Step 2: Find the critical value (from table I) (example: .123)

Critical r value is: 0.754

Critical t value 2.571

Step 3: Compute the test value using the formula or calculator (round to three decimal places, example .645):

r test value is:

t = \\frac{r} { \\sqrt{ \\frac{(1-r^2)}{n-2}}}

t = \\frac{0.9927} { \\sqrt{ \\frac{(1-0.9927^2)}{7-2}}}

t=18.404

calculate t=18.404 > 2.571, table value

Step 4: Reject the null or do not reject the null (type in either Reject the null or do not reject the null only): Reject the null

Step 5: Conclusion sentence (type in either isor is not only, to reflect what you found) and state if it is positive or negative.

There   is     enough evidence to support a   negative     relationship.

Find the equation of the regression line y\' = a + bx and fill in a and b below (round a and b to three decimal places, example: 4.123 or .234)

y\' =87.409 – 0.868 x

If there are 75 grams of water, how many grams of carbs? (do not round)

y\' =87.409 – 0.868*75

= 22.309

Regression Analysis

r²

0.9854

n

7

r

-0.9927

k

1

Std. Error

0.555

Dep. Var.

y

ANOVA table

Source

SS

df

MS

F

p-value

Regression

104.0074

1  

104.0074

337.98

8.76E-06

Residual

1.5387

5  

0.3077

Total

105.5461

6  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=5)

p-value

95% lower

95% upper

Intercept

87.4094

3.9099

22.356

3.33E-06

77.3586

97.4602

x

-0.8681

0.0472

-18.384

8.76E-06

-0.9894

-0.7467

Predicted values for: y

95% Confidence Interval

95% Prediction Interval

x

Predicted

lower

upper

lower

upper

Leverage

75

22.30551

21.22784

23.38317

20.51810

24.09291

0.571

Regression Analysis
	r²	0.9854	n	7
	r	-0.9927	k	1
	Std. Error	0.555	Dep. Var.	y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	104.0074	1	104.0074	337.98	8.76E-06
Residual	1.5387	5	0.3077
Total	105.5461	6
Regression output				confidence interval
variables	coefficients	std. error	t (df=5)	p-value	95% lower	95% upper
Intercept	87.4094	3.9099	22.356	3.33E-06	77.3586	97.4602
x	-0.8681	0.0472	-18.384	8.76E-06	-0.9894	-0.7467
Predicted values for: y
		95% Confidence Interval	95% Prediction Interval
x	Predicted	lower	upper	lower	upper	Leverage
75	22.30551	21.22784	23.38317	20.51810	24.09291	0.571