Question

Many fast-food restaurants have soft drink dispensers with preset amounts, so that when the operator merely pushes a button for the desired rink the cup is automatically filled. This method apparently saves time and seems to increase worker productivity. A researcher randomly selects 9 workers from a restaurant with automatic dispensers and 9 works from a restaurant with manual dispensers. At a 1% significance level, use the Mann-Whitney U Test to test whether workers with automatic dispensers are significantly more productive. Automatic (Group 1): 153, 128, 143, 110, 152, 168, 144, 137, 118 Manual (Group 2): 105, 118, 129, 114, 125, 117, 106, 92, 126 What is the alternative hypothesis for this study? There is no difference in worker productivity between restaurants with automatic and manual dispensers. • Worker productivity is higher with automatic dispensers. Worker productivity is lower with automatic dispensers. Automatic dispensers fill cups faster than manual dispensers. What rank will be given to the observation value, 118 that is in both the automatic and manual groups? (Round answer to 1 decimal). When rounding the U test statistic up to the next value, what is the p-value from the Mann Whitney Table of p-values? (Round to 4 decimal places) 0.0092 X What can be concluded from this study at a 1% significance level? Worker productivity is lower with automatic dispensers. Automatic dispensers fill cups faster than manual dispensers. • Worker productivity is higher with automatic dispensers. There is no difference in worker productivity between restaurants with automatic and manual dispensers.

Answer 1

The rank for 118 in both group will be 7.5

Explnation

Size in Ascending Order	Rank	Name of related sample A for sample1, B for sample2	Rank for A	Rank for B
92	1	B		1
105	2	B		2
106	3	B		3
110	4	A	4
114	5	B		5
117	6	B		6
118	7.5	B		7.5
118	7.5	A	7.5
125	9	B		9
126	10	B		10
128	11	A	11
129	12	B		12
137	13	A	13
143	14	A	14
144	15	A	15
152	16	A	16
153	17	A	17
168	18	A	18
Total			115.5	55.5

p-value = 0.0091

( use following R-code to get P-value)

> x=c(153,128,143,110,152,168,144,137,118)
> y=c(05,118,129,114,125,117,106,92,126)
> wilcox.test(x,y)

Wilcoxon rank sum test with continuity correction

data: x and y
W = 70.5, p-value = 0.009154
alternative hypothesis: true location shift is not equal to 0