Question

A dog food manufacturer is using a new process to make his dog food.  He believes the new process affects the taste of the food.  To test the new process, the company recruited 9 dogs. The dogs were given the original food and the time required to eat a standard portion of the food was measured. Two days later the same dogs were fed the same quantity of the new formula and the time was measured again. Eating faster is assumed to mean the taste was better.  Eating slower means the taste was worse The times are below. Test the hypothesis that the new process affects the taste of the dog food using an alpha level of .05.

Old Process	New Process
115	118
120	120
140	120
137	131
150	142
124	125
132	121
119	117
160	135

MAKE SURE TO ANSWER ALL OF THE QUESTIONS. SHOW YOUR WORK WHERE POSSIBLE.

a) What is the appropriate test?

b) State the null hypothesis (in words and with means).

c) State the alternative hypothesis (in words and with means).

d) Find the critical value(s) (explain how you found it).

e) Calculate the obtained statistic (show all of your work).

f) Report the results in proper format. Make a decision.

Answer 1

Old	new	diff= new-old	(d-dbar)^2
115 120 140 137 150 124 132 119 160	118 120 120 131 142 125 121 117 135	3 0 -20 -6 -8 1 -11 -2 -25 M: -7.56	111.42 57.09 154.86 2.42 0.2 73.2 11.86 30.86 304.31 S: 746.22

Since

A) we used paired t test

B) null and alternate hypothesis

H0:d=0.

( There is no differencevof means of new process and old process)

C)

Alternate hypothesis

H1:0

( There is differencevin means of new process and old process)

D) level of significance =0.05

Critical values = 2.306

E) test statistic

Mean: -7.56
μ = 0
S² = SS⁄df = 746.22/(9-1) = 93.28
S²_M = S²/N = 93.28/9 = 10.36
S_M = √S²_M = √10.36 = 3.22

T-value Calculation

t = (M - μ)/S_M = (-7.56 - 0)/3.22 = -2.35

F) decision

Since |t| = 2.35>2.306 so we reject the null hypothesis.

So there sufficient evidence to conclude that there is significant difference in means of old and new process .