Answer:
a) Null hypothesis:
Alternative hypothesis:
b) 
Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.
c) 
In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.
d) 
Here again, with the p-value being less than the significance level, we can reject the null hypothesis.
Step-by-step explanation:
1) Provided data and references
represents the average of the samples
denotes the standard deviation of the samples
indicates the number of samples
is the value we are examining
defines the significance level for the test.
t represents the specific statistic of interest
indicates the p-value relevant to the test (the variable of concern)
Define the null and alternative hypotheses.
To assess if the true mean is at least 10 hours, we must set up a hypothesis:
Part a
Null hypothesis:
Alternative hypothesis:
If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:
(1)
Part b
In this scenario t=-2.3, 
Initially, we need to calculate the degrees of freedom 
Since this is a left-tailed test, the p-value is determined by:
In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.
Part c
For this situation t=-1.8, 
We need to find the degrees of freedom
For the left-tailed test, the p-value is given by:
In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.
Part d
For this case t=-3.6,
Firstly, we find the degrees of freedom
Since we are conducting a left-tailed test, the p-value is calculated as:
Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.
