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
[tex]\bar{x}=257[/tex]
[tex]\mu = 257[/tex]
Since n>30
Standard deviation = [tex]\sigma = 41[/tex]
So,we will use z test
[tex]H_0:\mu \geq 261\\H_a:\mu < 261[/tex]
Formula : [tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the values
[tex]z=\frac{257-261}{\frac{41}{\sqrt{200}}}[/tex]
[tex]z=-1.37[/tex]
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 [tex]\mu \geq 261[/tex]