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.
One sample T hypothesis test:
μ : Mean of variable H_{0} : μ = 500 H_{A} : μ ≠ 500 Hypothesis test results:

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 ttest because my dataset is a sample, the tstat is 1.4539797. The Pvalue is .1485, and since this number is larger than the significance level, .05, we fail to reject the null.
Contingency table results:
Rows: Total Fat (g) Columns: Calories
ChiSquare test:
ChiSquare 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, chisquared, came out to 329.74982, in which the pvalue is less than .0001. Since the pvalue 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.
Analysis of Variance results:
Data stored in separate columns. Column statistics
ANOVA table

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 Fstat was 505.87406. The Pvalue 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.
One sample Z hypothesis test:
μ : Mean of variable H_{0} : μ = 500 H_{A} : μ ≠ 500 Standard deviation not specified. Hypothesis test results:

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 ztest, my data was extremely similar when compared to the ttest calculated with the same variable. While my tstat remained at 1.4539797, my Pvalue slightly changed from .1485 for the Ttest and .146 for my Ztest. Although numerically different, the Pvalue is still greater than the significance level of .05, and therefore we fail to reject the null.
Contingency table results:
Rows: Carbs (g) Columns: Calories
ChiSquare test:
ChiSquare 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, chisquared, came out to 130.33433, in which the pvalue is less than .0001. Since the pvalue 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.
Analysis of Variance results:
Data stored in separate columns. Column statistics
ANOVA table

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 Fstat was 473.54551. The Pvalue 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.
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Dec 13, 2017
The hypothesis of anova and t test should be about the population mean;
In result 2, wrong conclusion.