Statistical Test - Assignment Example

Statistical Test This section was used to perform preliminary statistical test on the data that were collected from who patronize the school cafeteria on their desire to be served with specific soft drinks and how they would be able to determine that the specific drink they demanded was what they were served with. The data were collected based on the hypothesis that the only way to determine the true brand of a product is through taste and not just by looking at the appearance of the product. The assertion was made against the backdrop that there could be imitation of packaging.

This section shall therefore deal with the organization, summarization, display, consideration and above all the evaluation of results. One major form of data are organized, summarized and displayed below. Inferential Statistics Chi Square Analysis Students’ specific preference for specific soft drinks Twenty students were asked if upon entry to the cafeteria, they made requests for specific soft drinks. The table below represents the responses that were received. Initial Data Table Student Yes No A.

Male with drinks only B. Male with a meal to drink C. Female with drinks only D. Female with a meal to drink 4 2 7 3 2 0 1 1 Summarized Data Table Student Yes No Total A B C D 4 2 7 3 1 0 2 1 5 2 9 4 Total 16 4 20 Expected Value Table Values in the expected table are calculated by a simple formula where cell value is multiplied by the row total and divided by the total table value. Student Yes No A B C D 0 0.2 3.15 0.6 0.25 0 0.9 0.2 Calculation of Chi Square X2 = Σ (O – E)2 E O= obtained frequencies E = expected value For cell A for instance the equation would be calculated as 4 – 0 42 = 16 16 ÷ 0 = 0 The same procedure results in the following values Chi Square Table Student Yes No Total A B C D 0 18 2.

44 8 6 0 2.44 8 6 18 4.88 16 TOTAL 28.44 16.44 44.88 Table total = 44.88 Evaluation of Chi Square The degrees of freedom (df) are the number of rows minus 1 times the number of columns minus 1 = (3 x 1) = 3df From Cookbook (p. 498), the probability for this value is > 0.05 This implies that 0.05 = 7.815 for 3df Meanwhile our present result is 44.88 This means that 44.88 > 7.815 This implies that the results are very significant. Evaluation of Results The results give a very detailed implication to the fact that the preference of people on the kind of drink they would want to be served with when they sit in any cafeteria cannot be underestimated.

Attendants are therefore admonished to make it a point to always inquire from customers, the kind of drinks they would want to be served with and also ensure that the specific results of customers are met. REFERENCE LIST Cookbook Approach to Statistical Analysis. (n.d). Unpublished.