Question
In: Statistics and Probability
A supermarket chain claims that its customers spend an average of 65.00 per visit to its...
A supermarket chain claims that its customers spend an average of 65.00 per visit to its stores. The manager of a local Long Beach store wants to know if the average amount spent at her location is the same. She takes a sample of 12 customers who shopped in the store over the weekend of March 18-19th. Here are the dollar amounts that the customers spent:
88
69
141
28
106
45
32
51
78
54
110
83
Calculate the mean and the standard deviation. Run the appropriate test in SPSS. You may choose to evaluate and make a decision based on p-value or critical value from the t-table.
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Answer:-
Given That:-
A supermarket chain claims that its customers spend an average of 65.00 per visit to its stores. The manager of a local Long Beach store wants to know if the average amount spent at her location is the same. She takes a sample of 12 customers who shopped in the store over the weekend of March 18-19th. Here are the dollar amounts that the customers spent:
88
69
141
28
106
45
32
51
78
54
110
83
Mean = 
Standard deviation 
t.test
degree
fraction
P - value is = 
we reject 
is p value <
level signification at 5% level of significance
.3920
.05
we can not reject
of means the null hypothesis is accepted .
orchestra answered 1 year agoRelated Solutions
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