Question

5)

Kellogg’s Cereal Corporation is performing a test of a new marketing strategy in 9 selected Midwest cities to see if replacing “Tony the Tiger” as the “face” of Kellogg’s Sugar Frosted Flakes will result in increased sales.

Data is in large boxes sold per store.

Market           Monthly sales before changing Tony         Sales the first month after replacing Tony.

Detroit                        240                                                                 266

Chicago          334                                                                 346     

Denver                        197                                                                 193

Houston          223                                                                 227

Oakland          134                                                                 151

Pittsburgh       231                                                                 236     

Sacramento     198                                                                 205

Toledo            101                                                                 119

Zanesville       123                                                                 138

Perform a t-test for Two Dependent Samples and call monthly sales before the change as Group 1.

1. At α = 0.05, what was your Critical Value?
2. What was the value of your Test Statistic?
3. Did you “Reject” or “Fail to Reject” the Null Hypothesis?
4. What was the p-value for this study?

Answer 1

Using the excel tool for a given dependent sample

Group1	Group2
240	266
334	346
197	193
223	227
134	151
231	236
198	205
101	119
123	138

The hypotheses are

Using excel the left tail t-test as

t-Test: Paired Two Sample for Means
	Group1	Group2
Mean	197.8889	209
Variance	5138.1111	4981
Observations	9	9
Hypothesized Mean Difference	0
df	8
t Stat	-3.6898
P(T<=t) one-tail	0.0031
t Critical one-tail	1.8595

a) t critical value at 0.05 level of significance =1.8595

b)The value of test statistic =-3.6898

c)Since P-value is less than the level of significance 0.05 hence we reject the null hypothesis and support the claim.

d) The P-value =0.0031