Ans:
n=45
sample mean=42.95
sample standard deviation=2.64
Test statistic:
t=(42.95-42)/(2.64/sqrt(45))
t=2.414
df=45-1=44
p-value(2 tailed)=tdist(2.414,44,2)=0.0200
As,p-value<0.1,we reject the null hypothesis
.There is sufficient evidence to conclude that the changes altered the customer rating.
rur the following problems, show all work AND clearly state your conclusion. Use the method that is requested for...
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