Answer:
We are supposed to find Is this sample result good evidence that the mean of all students in this town is less than 261
Sample size = n = 200


Since n>30
Standard deviation =

So,we will use z test

Formula :

Substitute the values


Now find the p value from the z table
P(z<-1.37)=0.0853
Let us suppose the significance level is 5 %
So, α = 0.05
p value > α
So, we accept the null hypothesis
