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?

1. Find the critical value (example .123)

Critical r value = ______

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

r test value is: ________

3. Reject the null or do not reject the null (type in either Reject the null or do not reject the null only):

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

There _____ enough evidence to support a _____ relationship.

5. 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\' = ______ + _______x

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

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

Solution

We have given that 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).

The hypothesis for the test is,

H₀ :   = 0 Vs H₁ :   0

First we calculate sample correlation coefficient (r).

This we can calculate by using EXCEL.

syntax : =correl(array 1,array 2)

where array 1 = water observations.

and array 2 = carbs observations.

This will gives us r =   - 0.99268

The test statistic is,

t = r * sqrt(n-2) / sqrt(1 - r²)   

the test statistic follows t distribution with n - 2 degrees of freedom.

where n = number of data pairs in the data.= 7

t = -0.99268 * sqrt(5) / sqrt(1 - (-0.99268)² )

= -18.384

take I t I = 18.384

Critical value we can find by using EXCEL.

alpha = 0.05

Syntax : tinv( probability , d.f.)

where probability = alpha = 0.05

d.f. = n - 2 = 5

critical value = 2.570

I t I > critical value

reject H₀ at 5% level of significance.

And conclude that there is a linear relation between water and carbohydrates.

There is enough evidence to support a linear relationship.

Find the equation of the regression line y\' = a + bx and fill in a and b below

The data is given below,

We have to find a and b.

where a is intercept or constant.

b is slope.

We can find by using EXCEL,

Syntax :

=slope( known y\'s , known x\'s )

where y = carbohydrates

x = water

This will gives us value of the slope (b) is - 0.868

And for the intercept the syntax is,

intercept (known y\'s , known x\'s)

This will gives us intercept or constant (a) = 87.409

The equation of the regression line ,

y\' = 87.409 + (-0.868) * x

slope is negative indicates that there is negative correlation between carbohydrates and water.

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

y\' = 87.409 + ( - 0.868) * 75 = 22.309

Hence the answer.

water(x)	carbs(y)
83.93	15.25
80.76	16.55
87.66	11.1
85.2	13.01
72.85	24.27
84.61	14.13
83.81	15.11