Answer to Question #182329 in Statistics and Probability for Princess C. Tingcang

Question

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)

Answer 1

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

$n=50, \mu=120, x=122.50, \sigma=6.50$

Using z-distribution statistics-

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

The tabulated value of z at $\alpha=0.05 1.645$

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