Question

The manager of a major retail store has taken a random sample of 25 customers. The average sale was $52.50. The population standard deviation is known to be $6.10. The manager would like to determine whether or not the mean sales by all customers are significantly more than $50. What is the p-value for the test

Answer 1

Answer:P value = 1 - 0.9793 = 0.0207

Explanation:

we can use Z value and normal distribution to find P value. P value is the area of beyond the value of Z value

sample mean (x.bar) = $52.20

Population mean (U) = $50

Sample Standard deviation (Sd) =$ 6.10

sample (n) = 25

Z =$\frac{(x.bar - U)}{Sd/\sqrt{n} }$ =$\frac{52.50 - 50}{6.10/\sqrt{25} }$

Z = 2.50/1.22 = 2.049280328 = 2.049

area (normal distribution table) = 0.9793

P value = 1 - 0.9793 = 0.0207