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1. Since food trucks have been gaining popularity, Mi Casa Front Porch is testing a seasonal food truck for special events around the Phoenix Metropolitan
1. Since food trucks have been gaining popularity, Mi Casa Front Porch is testing a seasonal food truck for special events around the Phoenix Metropolitan area. The data gathered by the research team identified that the average menu item from food trucks was $6.50 and daily revenue at the events was $9500.00 with a standard deviation of $250.00. The team gathered data from 20 days that the food truck was operating. To be profitable, the food truck would need to earn more than the average. Develop the hypothesis and interpret the test statistic at a .05 level of significance to whether or not Mi Casa Front Porch should continue with their seasonal food truck. 2. Answer the questions below in essay format. Your essay must include an introduction, a body, and a conclusion, and it must address all relevant parts of each question. Your response should be a minimum of 150 words in length and include both the appropriate statistics and support for the analysis and interpretation. Make sure to cite any source you use. Proper citation format for a source includes the name of the author(s), the title of the work, the date of the publication, and the page number if you directly quote the source. 3. Based on the data below, create and conduct the appropriate hypothesis test at a .05 significance level. Provide the hypothesis and interpretation of the results for the business scenario. Revenue $9,525.35 $9,633.55 $10,322.01 $15,011.50 $9,099.50 $10,342.22 $11,765.83 $8,325.40 $9,525.60 $10,622.60 $11,732.86 $13,455.55 $10,222.60 $9,626.40 $11,001.00 $15,235.65 $13,548.84 $10,249.71 $14,854.76 $9,131.22 Null Hypothesis (Ho): = $9500.00 Alternative Hypothesis (Ha): > $9500.00 Where = population daily revenue at the events Level of significance = 0.05 Upper critical value at 5% with n-1 = 20-1 = 19 degree of freedom = 1.7291 Decision Rule: reject Ho if test statistic value is bigger than upper critical value. Test statistic is one sample t test t = (Sample Mean - Population mean)/SE Here Sample Mean = 11161.61 Population Mean = 9500 SE = sd/sqrt(n) = 2134.698128/sqrt(20) = 477.3330 t = (11161.61-9500)/ 477.3330 = 3.4810
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