Answer:
We need to find the degrees of freedom given by:
Since is a right tailed test the p value would be:
If we compare the p value and the significance level given 
we see that 
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
We need to find the degrees of freedom given by:
Since is a right tailed test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.
