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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

Respuesta :

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

Z = 2.50/1.22 = 2.049280328 = 2.049

area (normal distribution table) = 0.9793

P value = 1 - 0.9793 = 0.0207

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