14.7.2: Practice on Nutrition
 Page ID
 19845
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
This second example will use the same formula, more or less. Instead of providing the standard deviations, you will need to calculate them. This example uses nutrition data from fast food restaurants from www.OpenIntro.org. You should probably try the previous practice problem because it gives more hints than this one will.
Scenario
When Dr. MO was on a lowcarb diet, she noticed that things that were low in carbohydrates tended to be high in fat, and vice versa. Now that she has the skills and the data set, we can test to see if there is a statistical relationship! Our analysis will look at the chicken dishes from one fast food restaurant using the nutritional data from OpenIntro.org.
Step 1: State the Hypotheses
What kind of linear relationship is it if it looks like as fat goes up, carbs go down?
Exercise \(\PageIndex{1}\)
What is a research hypothesis for this scenario?
 Answer

The research hypothesis should be: There is negative linear relationship between fat and carbohydrates in the sample of chicken dishes from one fast food restaurant.
Sometimes, people will add a description of what the negative linear relationship would look like to their research hypothesis: There is negative linear relationship between fat and carbohydrates in the sample of chicken dishes from one fast food restaurant such that foods with more fat tend to have fewer carbohydrates.
Let’s look at the scatterplot to see if it looks like a negative linear relationships (as one goes up, the other goes down), which should have a general linear trend from the top left to the bottom right.
Uhoh, already it looks like Dr. MO’s research hypothesis is not working out. There does seem to be a general linear trend, but it does not look like a negative correlation; it actually looks like a positive correlation!
Let’s state our null hypothesis so that we can get to the analysis to see what’s going on statistically.
Exercise \(\PageIndex{2}\)
What is a null hypothesis for this scenario?
 Answer

The null hypothesis should be: There is no linear relationship between fat and carbohydrates in the sample of chicken dishes from one fast food restaurant.
Before we get too far into the weeds, let’s consider why a correlational analysis is the appropriate analysis. Your first clue is the research hypothesis. We are not comparing whether the average fat content is higher than the average carbohydrate content; in other words, we aren’t looking at mean differences. Another clue is that we have means of two different variables, not means of two different groups. So that leads to recognizing that we have two quantitative variables, not a quantitative variable that we measured for two or more groups (qualitative variables). Whenever you’re not sure, check out the Decision Tree handout!
Step 2: Find the Critical Values
The critical values for correlations come from the Table of Critical Values of r again. We were not given any information about the level of significance at which we should test our hypothesis, so we will assume \(α = 0.05\) as always. From the table, we can see that we need the Degrees of Freedom, which is N–2. The data is provided below, (Table \(\PageIndex{1}\)) and shows that we have 13 chicken dishes that we are looking at. With p=.05 level, the critical value for N –2 (13–2=11) is critical rvalue is 0. 0.553.
Step 3: Calculate the Test Statistic
It’s easiest to calculate the standard deviation and Pearson’s r by using a table to do most of the calculations. If we sum each column in the table, we also can easily find the mean.
Item 
Total Fat 
Fat Difference: Fat Minus Mean 
Squared 
Total Carbs 
Carb Differences: Carb Minus Mean 
Squared 
Fat Diff * Carb Diff 

2 piece PrimeCut Chicken Tenders 
11 


19 



Chicken Tender 'n Cheese Slider 
12 


30 



Buffalo Chicken Slider 
13 


31 



3 piece PrimeCut Chicken Tenders 
17 


28 



Buttermilk Buffalo Chicken Sandwich 
24 


53 



Crispy Chicken Farmhouse Salad 
24 


26 



Buttermilk Crispy Chicken Sandwich 
26 


52 



5 piece PrimeCut Chicken Tenders 
28 


47 



Bourbon BBQ Chicken Sandwich 
31 


66 



Buttermilk Chicken Bacon & Swiss 
31 


56 



Buttermilk Chicken Cordon Bleu Sandwich 
35 


53 



Pecan Chicken Salad Sandwich 
44 


81 



Pecan Chicken Salad Flatbread 
46 


53 



SUM: 
\(\sum\) = 342 


\(\sum\) =595 



Let’s go through this table. The Item (first column) is what kind of chicken dish we are talking about. Although it’s not necessary to include this information, Dr. MO likes to show that the numbers in the second column (Total Fat) are connected to the numbers in the Total Carbs column because they are associated with the same dish. The data area ordered so that it goes from items with the least fat to items with the most fat. That means that the Total Carbs cannot ASLO be ordered from least to most carbs because they would lose their connection to the meal that they are associated with. We can put the data in any order that we’d like, but, for example, the PrimeCut Chicken Tenders must always have Total Fat of 11 and Total Carbs of 19.
Okay, so we talked about the two columns that have data in them now. The empty columns are what we need to fill with our calculations to finish both the formula for standard deviations:
\[s=\sqrt{\dfrac{\sum\left((X\overline {X})^{2}\right)}{N1}} \nonumber \]
and the formula for correlations:
\[ r = \cfrac{ \left( \cfrac{\sum ((x  \bar{X_x})*(y  \bar{X_y}) ) }{(N1)}\right) } {\left( \sqrt{\dfrac{\sum\left((x\overline {X_x})^{2}\right)}{N1}} \right) \times \left( \sqrt{\dfrac{\sum\left((y\overline {X_y})^{2}\right)}{N1}} \right)} \nonumber \]
First things first, though. If you look at the table, the first empty columns says “Fat Difference: Fat Minus Mean”. This means that we must find the mean for the Total Fat variable, then subtract that number from each of the scores for fat.
Exercise \(\PageIndex{3}\)
What is the average amount of fat for chicken dishes from this fastfood restaurant? What is the average amount of carbohydrates for chicken dishes from this fastfood restaurant?
 Answer

\[ \overline{X_F} = \dfrac{\sum X}{N} = \dfrac{342}{13} = 26.31 \nonumber \]
\[ \overline {X_C} = \dfrac{\sum X}{N} = \dfrac{595}{13} = 45.77 \nonumber \]
Hopefully these data makes it a little more clear why we are more interested in finding a linear relationship between these two variables, rather than seeing if the means are different. We could conduct a ttest to see if the means are different, but what question would that answer? Dr. MO’s prediction was that items with more carbs had less fat, so finding that chicken dishes have more carbohydrates than fat doesn’t help us answer that question at all.
Now that we have the means for our two quantitative variables, we can start filling in the table.
Example \(\PageIndex{1}\)
Fill in Table \(\PageIndex{1}\) with the differences for each variable, square those differences, and multiply those differences (not the squares, though). At the end, you should have each column summed.
Solution
Item 
Total Fat 
Fat Difference: Fat Minus Mean 
Squared 
Total Carbs 
Carb Differences: Carb Minus Mean 
Squared 
Fat Diff * Carb Diff 

2 piece PrimeCut Chicken Tenders 
11 
15.31 
234.40 
19 
26.77 
716.63 
409.85 
Chicken Tender 'n Cheese Slider 
12 
14.31 
204.78 
30 
15.77 
248.69 
225.67 
Buffalo Chicken Slider 
13 
13.31 
177.16 
31 
14.77 
218.15 
196.59 
3 piece PrimeCut Chicken Tenders 
17 
9.31 
86.68 
28 
17.77 
315.77 
165.44 
Buttermilk Buffalo Chicken Sandwich 
24 
2.31 
5.34 
53 
7.23 
52.27 
16.70 
Crispy Chicken Farmhouse Salad 
24 
2.31 
5.34 
26 
19.77 
390.85 
45.67 
Buttermilk Crispy Chicken Sandwich 
26 
0.31 
0.10 
52 
6.23 
38.81 
1.93 
5 piece PrimeCut Chicken Tenders 
28 
1.69 
2.86 
47 
1.23 
1.51 
2.08 
Bourbon BBQ Chicken Sandwich 
31 
4.69 
22.00 
66 
20.23 
409.25 
94.88 
Buttermilk Chicken Bacon & Swiss 
31 
4.69 
22.00 
56 
10.23 
104.65 
47.98 
Buttermilk Chicken Cordon Bleu Sandwich 
35 
8.69 
75.52 
53 
7.23 
52.27 
62.83 
Pecan Chicken Salad Sandwich 
44 
17.69 
312.94 
81 
35.23 
1241.15 
623.22 
Pecan Chicken Salad Flatbread 
46 
19.69 
387.70 
53 
7.23 
52.27 
142.36 
SUM: 
\(\sum\) = 342 
\(\sum\) = 0.03 
\(\sum\) = 1536.77 
\(\sum\) = 595 
\(\sum\) = 0.01 
\(\sum\) = 3842.31 
\(\sum\) = 1997.92 
What you should see when you look at the table is that the sum of the differences from the mean are really close to zero, which means that we probably the calculations correctly.
From here, there are two different options that you can take. One option is to calculate the standard deviations separately using the same formula that we’ve used before:
\[s=\sqrt{\dfrac{\sum\left((X\overline {X})^{2}\right)}{N1}} \nonumber \]
Then add those standard deviation results into the formula that we used in our mental health example.
\[ r= \cfrac{ \left( \cfrac{\sum ((x_{Each}  \bar{X_x})*(y_{Each}  \bar{X_y}) ) }{(N1)}\right) } {(s_x \times s_y)} \nonumber \]
It is totally fine if you choose this option. At this point, it’s whatever you feel more comfortable with. If you choose this option, meet us at the end to make sure that you got the same answer!
What the example is going to do is the second option, and that is to fill in this crazy formula with all of the numbers from the table:
\[ r = \cfrac{ \left( \cfrac{\sum ((x  \bar{X_x})*(y  \bar{X_y}) ) }{(N1)}\right) } {\left( \sqrt{\dfrac{\sum\left((x\overline {X_x})^{2}\right)}{N1}} \right) \times \left( \sqrt{\dfrac{\sum\left((y\overline {X_y})^{2}\right)}{N1}} \right)} \nonumber \]
Hopefully you can see that they are the same formula, but replacing the standard deviation (s) of each variable with the formula for standard deviations.
This is going to be a doozy!
Example \(\PageIndex{2}\)
Fill in the sums from Table 2 to complete the Pearson’s correlation formula:
\[ r = \cfrac{ \left( \cfrac{\sum ((x  \bar{X_x})*(y  \bar{X_y}) ) }{(N1)}\right) } {\left( \sqrt{\dfrac{\sum\left((x\overline {X_x})^{2}\right)}{N1}} \right) \times \left( \sqrt{\dfrac{\sum\left((y\overline {X_y})^{2}\right)}{N1}} \right)} \nonumber \]
Solution
First, Dr. MO likes to replace the "x" and "y" with F for Fat and C for Carbs. That helps her follow the formula better. That gives us:
\[ r = \cfrac{ \left( \cfrac{\sum ((X_F  \bar{X_F})*(X_C \bar{X_C}) ) }{(N1)}\right) } {\left( \sqrt{\dfrac{\sum\left((X_F\overline {X_F})^{2}\right)}{N1} } \right) \times \left( \sqrt{\dfrac{\sum\left((X_C\overline {X_C})^{2}\right)}{N1}} \right)} \nonumber \]
As long as you follow the order of operations, you can fill in the formula however you’d like. You can start with the numerator, and ignore the denominator for a bit, or fill in the two standard deviations first, then fill out the numerator. Again, it’s whatever makes the most sense to you. Dr. MO likes to plug in all of the numbers at once, then do sets of parentheses to start making the formula smaller. So here it is with all of the numbers:
\[ r = \cfrac{ \left( \cfrac{1997.92 }{(131)}\right) } {\left( \sqrt{\dfrac{1539.77}{131}} \right) \times \left( \sqrt{\dfrac{3842.31}{131}} \right)} \nonumber \]
Remember for standard deviations that you square each difference score, then add up all of the squares (sum of squares!). We did that in the table, so we just plug those numbers in. We also already got the entire top part of the numerator from the table.
Next step is to start calculating what’s in the parentheses. This first set will be the N1’s, then the first set of division.
\[ r_{N1's} = \cfrac{ \left( \cfrac{1997.92 }{12}\right) } {\left( \sqrt{\dfrac{1539.77}{12}} \right) \times \left( \sqrt{\dfrac{3842.31}{12 }} \right)} \nonumber \]
\[ r_{Division} = \cfrac{ 166.49 } {\left( \sqrt{128.06} \right) \times \left( \sqrt{320.19} \right)} \nonumber \]
A couple more steps to get down to one number!
\[ r_{SquareRoot } = \dfrac{ 166.49 } {\left( 11.32 \right) \times \left( 17.89 \right)} \nonumber \]
\[ r_{Multiply} = \dfrac{ 166.49 } {\left( 202.49 \right)} \nonumber \]
\[ r = \dfrac{ 166.49 } { 202.49 } = 0.82 \nonumber \]
So our calculated correlation between fat and cars is \(r = 0.82\), which is a strong, positive correlation. Now we need to compare it to our critical value to see if it is also statistically significant.
Step 4: Make a Decision
Our critical value was is r_{Critical} = 0. 0.553 and our calculated value was \(r_{Calc} = 0.82\). Our calculated value was larger than our critical value, so we can reject the null hypothesis because this is still true:
Critical \(<\) Calculated \(=\) Reject null \(=\) There is a linear relationship. \(= p<.05 \)
Critical \(>\) Calculated \(=\) Retain null \(=\) There is not a linear relationship. \(= p>.05\)
What does the statistical sentence look like?
Exercise \(\PageIndex{4}\)
What is the statistical sentence for the results for our correlational analysis?
 Answer

The statistical sentence is: r(11)=0.82, p<.05
Okay, we're ready for the final writeup! Because our r=0.82, a positive number, the linear relationship is also positive. (Hint: This is not what Dr. MO hypothesized, but it does match the scatter plot!)
Example \(\PageIndex{3}\)
Write up your conclusion. Don't forget to include the four requirements for reporting results!
Solution
"We hypothesized that there would be a negative linear relationship between fat scores (M=26.31) and carbs scores (M=45.77) such that the chicken meals with more fat would have less carbs. Although there is a strong correlation between fat and carbs (r(11)=0.82, p<.05), the relationship is the opposite of the research hypothesis; chicken meals with more fat also tended to have more carbs."
And that’s how you calculate and interpret a Pearson’s correlation when you’re given the raw data!
Next, we will learn about some options when we don’t think that our distributions are normally distributed.
Contributors and Attributions
Foster et al. (University of MissouriSt. Louis, Rice University, & University of Houston, Downtown Campus)