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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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 .


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