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A restaurant's manager is interested
in testing the mean daily sales of his
restaurant. Assume that the
population is normally distributed
with a standard deviation of $250.
Over the first 20 days, the mean
daily sales were $6100. The manager
claims that the daily average sales
are greater than 6000 using a level
of significance of 0.01. The null
hypothesis is: *

Respuesta :

mysticchacha

Answer:

Null Hypothesis : Test true mean  u  > 6000

However our p-value turned out to be 0.0375

Cannot reject null hypothesis.

Step-by-step explanation:

You want to assume the true mean  u  > 6000

We test this hypothesis.

s = 250

average m = 6100  with sample size n = 20

P ( u > 6000)

P ( (u - m / s )* root(n)  > ((6000 -6100)/250)* root(20) )  ?

P ( (u - m / s )* root(n)  > 1.7888 )  = 0.0375

at the 96.25% confidence interval   But 0.01 means that it is out of the Confidence Interval, Cannot reject this number.

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