In: Statistics and Probability
The working assumption in the company is that customers spend on average $300 on the company's services. Recently, you started to suspect that things maybe not going to well and that your customers are not spending as much as they did. Set the null and alternative hypotheses that would reflect your concern and then test them at the 5% level of significance using the data in the After_Class_Assignment_Data Excel file.
Observation | Average Annual Spending | Year of First Trsnaction |
Customer 1 | $392 | 2014 |
Customer 2 | $57 | 2015 |
Customer 3 | $297 | 2013 |
Customer 4 | $329 | 2014 |
Customer 5 | $361 | 2016 |
Customer 6 | $258 | 2016 |
Customer 7 | $351 | 2016 |
Customer 8 | $367 | 2010 |
Customer 9 | $197 | 2017 |
Customer 10 | $450 | 2013 |
Customer 11 | $94 | 2017 |
Customer 12 | $105 | 2017 |
Customer 13 | $68 | 2010 |
Customer 14 | $293 | 2017 |
Customer 15 | $75 | 2012 |
Customer 16 | $172 | 2010 |
Customer 17 | $75 | 2010 |
Customer 18 | $290 | 2011 |
Customer 19 | $282 | 2011 |
Customer 20 | $434 | 2010 |
Customer 21 | $277 | 2013 |
Customer 22 | $142 | 2010 |
Customer 23 | $366 | 2015 |
Customer 24 | $464 | 2012 |
Customer 25 | $216 | 2013 |
since,you started to suspect that things maybe not going to well
so, concern is regarding that comapny may spend less than $300 on average.
Ho : µ = 300
Ha : µ < 300
sample std dev , s = 129.2359
Sample Size , n = 25
Sample Mean, x̅ = 256.480
degree of freedom= DF=n-1=
24
Standard Error , SE = s/√n =
25.8472
t-test statistic= (x̅ - µ )/SE =
-1.684
p-Value = 0.0526 [excel formula
=t.inv(-1.68,24) ] ]
Conclusion: p-value>α=0.05, Do not reject
null hypothesis
so, there is no enough evidence that to conlcude that customers spend on average less than $300 on the company's services at α=0.05