Hello, i need someone to check my response to my scenario. someone said this is slightly incorrect. im not sure if that is true. The hypothesis...

i need someone to check my response to my scenario. someone said this is slightly incorrect. im not sure if that is true.

The hypothesis scenario:

A doctor wants to know if the median BMI of a group of 100 patients with presumed overweight is equal to 30. (this is the medical value to know if you are overweight) Given the sample mean of 31.

Based on historical data, the doctor knows that these patients have a standard deviation of 5, so he uses this value as the standard deviation of the population in a Z test of 1 sample.

My response:

Null hypothesis H0: μ =30

Alternative Hypothesis H1; μ >30

You can put this in calculator

z0​=5/100​ √31−30​=2

=1-Norm.dist(2,0,1 true) ???

=0.02275

=Norm.s.inv ____?

Using the Z to P value calculator with α =0.05 and z=2, you get 0.9772

P value= 1−0.9772=0.0228

Mo : le = 30H, : M 7 3 D7 = 31T = 5n = 100Z = Z-l31-305= 2p - value = P(z xz )= P ( Z72 )I excel function= 1 - NounDist ( 2, 0, 1, truel ) )= 0 . 02275d = 0.05Sinceprvalue is less...