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Created: Dec 13, 2017
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Nutritional Data for Fast Food 2017 Report Part 4
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Data set 1. Nutritional Data for Fast Food 2017   [Info]
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 My dataset is a sample because it only surveys a portion of fast food restaurants rather than an entire population such as all of the fast food restaurants in the whole country or world.


Result 1: One sample T hypothesis test: calories   [Info]
One sample T hypothesis test:
μ : Mean of variable
H0 : μ = 500
HA : μ ≠ 500

Hypothesis test results:
VariableSample MeanStd. Err.DFT-StatP-value
Calories532.4920622.3469861251.45397970.1485

 

The variable used in the test was calories. The null hypothesis (Ho) was sample mean=500, while the alternative hypothesis (Ha) was the sample mean ≠ 500. After performing a t-test because my dataset is a sample, the t-stat is 1.4539797. The P-value is .1485, and since this number is larger than the significance level, .05, we fail to reject the null.


 

Result 2: Contingency table (with data)   [Info]
Contingency table results:
Rows: Total Fat (g)
Columns: Calories
0 to 200200 to 400400 to 600600 to 800800 to 10001000 to 12001200 to 1400Total
0 to 2073411100053
20 to 4007231010041
40 to 6000210100022
60 to 8000002709
80 to 10000000011
Total74136211371126

Chi-Square test:
StatisticDFValueP-value
Chi-square24329.74982<0.0001
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

 

The variables included in the contingency tables are calories and total fat (g). While the null hypothesis states that there is no relationship between calories and total fat, the alternative hypothesis states that the two factors are dependant and maintain a relationship. The statistic, chi-squared, came out to 329.74982, in which the p-value is less than .0001. Since the p-value is less than the significance level, I reject the null hypothesis. Therefore, it can be concluded that there is no relationship between calories and total fat.


 

Result 3: One Way ANOVA   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Total Fat (g)12628.54444418.2409631.6250341
Calories126532.49206250.8442922.346986

ANOVA table
SourceDFSSMSF-StatP-value
Columns11599968215999682505.87406<0.0001
Error2507906949.131627.796
Total25123906631

 

The variables included in the One Way ANOVA are calories and total fat (g). The null hypothesis is that the sample mean of calories and total fat are equal while the alternative hypothesis is that the samples means of both calories and total fat are not equal. Using the ANOVA statistic test, it was found that the F-stat was 505.87406. The P-value was less than .0001, and since this is less than the significance level of .05, the null hypothesis can be rejected, indicating that the sample means are not equal.


Result 4: Extra Credit: One sample Z hypothesis test   [Info]
One sample Z hypothesis test:
μ : Mean of variable
H0 : μ = 500
HA : μ ≠ 500
Standard deviation not specified.

Hypothesis test results:
VariablenSample MeanStd. Err.Z-StatP-value
Calories126532.4920622.3469861.45397970.146


Extra credit #1: The variable used in the test was calories. The null hypothesis (Ho) was sample mean=500, while the alternative hypothesis (Ha) was the sample mean ≠ 500. After performing a z-test, my data was extremely similar when compared to the t-test calculated with the same variable. While my t-stat remained at 1.4539797, my P-value slightly changed from .1485 for the T-test  and .146 for my Z-test. Although numerically different, the P-value is still greater than the significance level of .05, and therefore we fail to reject the null.


Result 5: Extra Credit #2: Contingency table (with data)   [Info]
Contingency table results:
Rows: Carbs (g)
Columns: Calories
0 to 200200 to 400400 to 600600 to 800800 to 10001000 to 12001200 to 1400Total
0 to 20751000013
20 to 4002810100039
40 to 60081413106152
60 to 80006321012
80 to 10000330006
100 to 12000211004
Total74136211371126

Chi-Square test:
StatisticDFValueP-value
Chi-square30130.33433<0.0001
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.


Extra credit #2: The variables included in the contingency tables are calories and Carbs (g). While the null hypothesis states that there is no relationship between calories and carbs, the alternative hypothesis states that the two factors are dependant and maintain a relationship. The statistic, chi-squared, came out to 130.33433, in which the p-value is less than .0001. Since the p-value is less than the significance level (.05), I reject the null hypothesis. Therefore, it can be concluded that there is a relationship between calories and total fat.


 

Result 6: Extra Credit #3: One Way ANOVA   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Calories126532.49206250.8442922.346986
Carbs (g)12644.57222220.5227351.8283106

ANOVA table
SourceDFSSMSF-StatP-value
Columns11499814414998144473.54551<0.0001
Error2507918005.331672.021
Total25122916149


Extra credit #3: The variables included in the One Way ANOVA are calories and carbs(g). The null hypothesis is that the sample means of calories and carbs are equal to one another while the alternative hypothesis is that the samples means of both calories and carbs are not equal to one another. Using the ANOVA statistic test, it was found that the F-stat was 473.54551. The P-value was less than .0001, and since this is less than the significance level of .05, the null hypothesis is rejected, indicating that the sample means are not equal.


 

Extra Credit #4: While my dataset contains enough variables and a big enough sample to sufficiently make claims and draw conclusions regarding the nutritional data for fast food 2017, more restaurants, in addition to a wider array of both items and types of foods could have been surveyed to ensure even more accurate and precise data collection. In addition, other factors could have impacted the nutritional data of the foods, such as what farms or places the meat was purchased from or how it is cooked.

 

HTML link:
<A href="https://www.statcrunch.com/5.0/viewreport.php?reportid=74587">Nutritional Data for Fast Food 2017 Report Part 4</A>

Comments
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By xg15
Dec 13, 2017

The hypothesis of anova and t test should be about the population mean;
In result 2, wrong conclusion.

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