i am having a difficult time with the norm.dist and norm.inv they are not properly formated and in order.

they are not properly formated and in order. im not sure if my caculation and my explanation is correct for the dist. thank you for checking my work i created.

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 explanation:

Null hypothesis H0: μ =30

Alternative Hypothesis H1; μ >30

You can put this in excel or your calculator

(30-31)/(100-5)=2

or enter this in excel

=Norm.dist(30-31)/(100-5)2=,true)

Then take that value which we will call A

and put this in excel

=Norm.s.inv

Using the z to p value calculator with a=0.05 and z=2, you get 0.9772

P value= 1−0.9772=0.0228

my Conclusion:

Since the P-value is less than the value of α , we REJECT the null hypothesis. There is not sufficient evidence to support the claim that the median BMI is 30.