Question #182329

the cashier of a fastfood restaurant claims that the average amount spent by customers for dinner is P120. A sample of 50 customers over a month period was randomly selected and it was found out that the average amount spent for dinner was P122.50. Using a 0.05 level of significance, can it be concluded that the average amount spent by customer was more than P120? Assume that the population standard deviation is P6.50. (Use the low-value method)


1
Expert's answer
2021-05-07T12:05:47-0400

Let HoH_o : It can be concluded that the average amount spent by customer is 120


n=50,μ=120,x=122.50,σ=6.50n=50, \mu=120, x=122.50, \sigma=6.50


Using z-distribution statistics-


z=(xμ)σ=(122.5120)6.5=0.384z=\dfrac{(x-\mu)}{\sigma}=\dfrac{(122.5-120)}{6.5}=0.384



The tabulated value of z at α=0.051.645\alpha=0.05 1.645


Conclusion: The calculated value of z is less than the tabulated value so HoH_o is accepted, i.e. It can be concluded that the average amount spent by costumer was 120.


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