Question

Cris Turlock owns and manages a small business in San Francisco, California. The business provides breakfast and brunch food, via carts parked along sidewalks, to people in the business district of the city. Being an experienced businessperson, Cris provides incentives for the four salespeople operating the food carts. This year, she plans to offer monetary bonuses to her salespeople based on their individual mean daily sales. Below is a chart giving a summary of the information that Cris has to work with. (In the chart, a sample" is a collection of daily sales figures, in dollars, from this past year for a particular salesperson) Groups Sample Sample variance Salesperson 1 Salesperson 2 Salesperson 3 Salesperson 4 123 95 101 115 Sample mean 206.7 199.0 216.3 222.5 2619.7 2916.9 34336 2483.2 Cris' first step is to decide if there are any significant differences in the mean daily sales of her salespeople. (If there are no significant differences, she'll split the bonus equally among the four of them.) To make this decision, Cris wil do a one-way, Independent samples ANOVA test of equality of the population means, which uses the statistic Variation between the samples Variation within the samples For these samples, F3.99. 0 Give the p-value corresponding to this value of the F statistic, Round your answer to at least three decimal places Can we condude, using the 0.01 level of significance, that at least one of the salespeople's mean daily sales is significantly different from that of the others? Yes NO X 5 ?

Answer 1

Answer:

Since in the above question given that the F statistic is 3.99

F = 3.99

P- value = 0.0080

Since the degree of freedom of numerator = 4-1= 3

Degree of denominator = n-k = 434-4= 430

Because n = 434 and k = 4

since the p- value is less than level of significance i.e. 0.0080 < 0.01 so we reject the null hypothesis.

So yes, we can conclude that at least one of the salespeople mean daily sales is significantly different from others.