I am having a difficult time setting my format like the student example instructions for instance my norm.dist and norm.s.inv does not look like the student example format. im not sure if my format is correct and my explanation is corect like the student example. 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: =30

Alternative Hypothesis H1; >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= 10.9772=0.0228

my Conclusion:

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

student example instructions:

Then make your own hypothesis test.For instance, XYZ car company boast that it's new car Eco Auto gets at least 59 miles to the gallon. Given a sample mean of 57 and a standard deviation of 3.5 where 35 people were tested, find the z value and p score. Hint: Use the P value calculator in the announcements to find the p score. Note: On this one you do not need to attach an excel spreadsheet.

Student Response:

Ho: u>=59

Ha: u<59

You can put this in excel or your calculator

= (59-57)/(3.5/35^.5)

=3.38

or enter this in excel

=norm.dist(59,57,3.5/35^.5,true)

and hit enter

Then take that value which we will call A

and put this in excel

=norm.s.inv(A)

and hit enter

Using the z to p value calculator with a=.05, 1 tailed test and z=3.38, you get

p=.000362

Since p < .05, you reject the null