Answer:
Given : The sample mean score is 51.48 .
The standard deviation for the population of test scores is 15.
The nationwide average score on this test is 50.
To Find : Perform the hypothesis test and compute the P-value?
Solution:
Sample size = n = 64
Since n > 30
So, we will use z test over here
We are supposed to find The school superintendent wants to know whether the second-graders in her school district have greater math skills than the nationwide average.
Hypothesis : 
Formula of z test : 

Substitute the values


Find the value of P(z) from z table
p(z)=0.7823
Since we are given that we are supposed to find The school superintendent wants to know whether the second-graders in her school district have greater math skills than the nationwide average.
So, p value = 1-P(z) = 1-0.7823=0.2177
In general , Level of significance is 0.05
So, p value is greater than alpha.
So, we accept the null hypothesis.