Water and Carbohydrates. Continuing on the theme of fruits and vegetables, here
ID: 3390914 • Letter: W
Question
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?
Ho: ? = 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-0.9926Explanation / Answer
Ho: = 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 = rac{r} { sqrt{ rac{(1-r^2)}{n-2}}}
t = rac{0.9927} { sqrt{ rac{(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
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