Mr. Goodbar Nuts Observational Study The Mr. Goodbar candy bars consist of chocolate and nuts. Some people enjoy having a regular size, 52g, while on the other hand others prefer the king size, 96g, Mr. Goodbar candy bar as a snack. In my study I wanted to see what percent of Mr. Goodbar candy bars are nuts. In addition, I wanted to find out how the size of the candy bar would affect the amount of nuts. The population that we sampled was all Mr. Goodbar candy bars weighing both 52g and 96g. The sampling was done by randomly selecting the stores in towns of my choice. Below I have included the store that I picked the candy bars from the Salem/Keizer area: Walgreens, Safeway, Wal-Mart, 7/11, Fredmyers, trader Joes. As I went to a different …show more content…
As well as prove the point that the standard deviation shows that the data was all close together. The test of significance is designed to assess the strength of the data against the null hypothesis. A test of significance assesses this in terms of probability. In this study our null hypothesis or H0 states that the percentage of nuts for all 52g candy bars is equal to the percentage for all 96g candy bars. The other statement being tested in a test of significance is called the alternative hypothesis or Ha. In our study this statement states that the percentage of nuts in 52g candy bars does not equal the percentage of nuts in 96g candy bars. The H0 is proven to be true as our P-Value of .72. The P-Value is the probability that the test statistic would take a value as extreme or more extreme than that actually observed, assuming that H0 is true. The larger the P-Value is, the stronger the evidence to support H0 provided by the data. This information would be of interest to the Mr. Goodbar manufacturing company, due to they should be informed in the percentage of nuts in each Mr. Goodbar. People that buy this product should be interested in the percentage of nuts if they have a preference in their candy bar being either mainly nuts or chocolate. For example if an individual dislikes nut then this is not the candy bar for them to invest
Within the target site of the experiment, researchers wanted to answer their hypothesis; hypothesis was that increased police
5. Andrew has $20 to purchase candy bars. If each candy bar cost $1.50, how many candy bars can Andrew purchase. With c representing the number of candy bars, write and evaluate the expression.
During Valentine’s week alone, millions of pounds of chocolate candies alone are sold (“Who consumes the most chocolate,” 2012, para 8). This naturally creates a demand for product, which in turns causes a need for ingredients. The main component in chocolate, of course, is cocoa. Since Côte d’Ivoire provides 40 percent of the world’s supply of this crucial ingredient (Losch, 2002, p. 206), it merits investigation i...
For this experiment the null hypothesis is that the intensity of the step rate test (High and Low) has no effect on the persons’ heart rate and recovery time. While the alternate hypothesis is that the intensity of the step rate test (High and Low) has an effect on the persons’ heart rate and recovery time.
The overall average of the control Daphnia’s heart rate is 249.38 bpm. 0.01% caffeine’s average is 327.93 bpm, and the caffeine at 0.005% has an average of 268.90 bpm, both making the heart rate speed up. Ethanol had the opposite affect, 0.01% ethanol’s average heart rate for this experiment is 159.58 bpm and 0.005% ethanol had an average of 183.4 bpm. Caffeine has a positive percent change while ethanol has a negative percent change in the data chart. The percent change for 0.01% caffeine is 31.50%change, for 0.005% of caffeine it is 7.83% change and for ethanol 0.01% it is 36.01% change while 0.005% ethanol has a 26.47% change. The standard deviation for the treatments all relatively close. Caffeine 0.01% had a standard deviation of 49.77, 0.005% caffeine’s standard deviation is 58.95. The standard deviation for 0.01% ethanol is 54.19, ethanol 0.005 had a standard deviation of 49.47, and the control groups is 33.31. The p-tests show if and how significant the data
In this analysis the null hypothesis is that the variable are independent, in other words whether or not a person has gotten their flu shot is unrelated to which group they are in. The Alternate hypothesis is that whether or not a person has gotten their flu shot is related to which group they were placed in, the variables are not independent. The results of this analysis are that the chi-square value is 4.1620049, which is nonsignificant according to the table on page 416 of the text which shows that the level of significance for 2 degrees of freedom is 5.99. The p value of 0.1248 is also indicative of a nonsignificant result. Based on the results of this analysis and the resulting significance the keep the null hypothesis.
...will fall within the first standard deviation, 95% within the first two standard deviations, and 99.7% will fall within the first three standard deviations of the mean. The Empirical Rule is used in statistics for showing final outcomes. After a standard deviation is found, and before exact data can be collected, this rule can be used as an estimate to the outcome of the new data. This probability can be used for gathering data that may be time consuming, or even impossible to found. When the mean equals the median and the values cluster around the mean and median, producing a bell-shaped distribution, then we can use the empirical rule to examine the variability. In this bell-shaped data set, we can calculate the mean and the standard deviation. The mean means the average value of the set of data. The standard deviation means the average scatter around the mean.
The inquiry science lesson that was chosen was Candy Heart Science Observations. Students were asked to determine if candy hearts will sink or float. Students were also asked to
Also, the title of the article states the research is a “population study” which is a focus of a quantitative research and a component of a quantitative method. Furthermore, the authors specified a clear defined research purpose which often requires statistical methods to test the hypotheses as well as to look for the cause and effects of the variables so that predictions can be
The null hypothesis being tested states that there is no difference in maple tree density in rural versus urban habitats. The experiment conducted observed maple tree density in both rural and urban habitats. The observed mean number of maple trees per quadrant in an urban habitat (Jackson Park) was 2 trees. The observed mean number of maple trees per quadrant in a rural habitat (Ojibway) was 6.25 trees. A student’s t-test was performed to test the null hypothesis. The tree density was measured in 16 quadrants in a total from the two habitats, thus the degrees of freedom of the experiment was 14. The critical t-value was 2.14 and the observed t-value was 3.12. Since the observed t-value is greater than the critical t-value, the results are
Chocolate bars are thought of as impulse buys, which means they require no thought. This is due to how inexpensive they are. However, if an ingredient such as sugar was to rise drastically, so will the cost of the chocolate bar therefore changing the buyer's perspective on the product class.
The recent product, liquor filled chocolates, is a viable business that can sell if it is implemented professionally. This recent innovation should be able to acquire attention from the market owing to its combination of selling products. Put simply, the liquor-filled chocolates are chocolates that contain alcohol. According to Novellino (2011), chocolate-candy sales summed up to $16 billion in 2008 in the U.S. Furthermore, the statistics on alcohol reveals that liquor sales hit $19.9 billion in 2011.
Recently the company sales was hit with a growing demand for low-carb snack bars. Customer preference has changed towards the NRG-A and NRG-B bars and so they want a product with low-carbohydrates in it. Fitter Snacker decides to put a new low-carb bars into the market because of its plans to remain in competition even though it isn’t recording any lost in sales.
Chi-Square is a statistical test that is utilized to make comparisons of observed data with data that the researcher expects to find with respect to a specified hypothesis. The test is used to determine whether the deviations in the data observed from the expected data have occurred just by chance or is caused by other factors (Brooks, 2008). The Chi-Square is usually employed to test the null hypothesis. For instance, it can be used to test whether there is no significant difference between the expected and observed outcomes.
The U.S. doughnut industry was a $5 to $6 billion market in 2003 2004 with a robust growth rate of 13%. Doughnut specialty stores were the fastest growing dining category in 2002 2003 with sales increases of 9% to approximately $3.6 billion. Opportunity for expansion in North America and globally is desirable. Doughnuts appeal to many people across all ages and demographics. The increasing rate of obesity and the concern about healthy living triggers a change in buyer demand toward a more health conscious diet.