Computation Involving Average Claim, Random Sample and Mean, and Alpha (Math Problem Sample)

Please read each problem carefully. Write your answers legibly. Show all your work. State all the necessary work that you would provide for a hypothesis testing situation. Answer all questions in the space provided. Each problem is worth 32 points.
1 A school transportation manager claims that her drivers spend an average of 36 hours in professional development training per school semester, with a population standard deviation of σ = 16 hours. A survey of a random sample of n = 36 drivers revealed a sample mean time of 40 hours of professional development training. Using an alpha of .05, determine if the amount of time spent in professional development training is significantly different from the claim of 36 hours. Assume that the data are normally distributed. Does the transportation manager have a valid claim? Provide proper interpretation.
average claim = 36
population sd 16hrs
random sample = 36
sample mean = 40hrs
alpha = 0.05H0: Mu = 36 vs the H1: Mu not equal to 36
critical values -1,96 and 1,96
the z value = (40-36)/(16/sqrt36)
= 1.5
The transport manager have a claim.
Our calculated z value is below the critical value of 1.96 therefore we accept the null hypothesis that drivers spend an average of 36hrs in professional development training per school semester and reject the alternative hypothesis.
2 The Management team of a certain company argued that its health care workers were getting paid an average hourly wage of $18.98. The union representative for the health care workers argues that the average hourly pay is lower. A sample of 25 employees revealed an average hourly pay of $18.10 with a sample standard deviation of $6.15. Using an alpha of .01, determine if the average hourly pay is significantly lower than the management claim. Assume that data are normally distributed. Does the Management team have a valid claim? Or do the workers have a valid claim? Provide proper interpretation.
mean 18.98
sample n 25
average 18.10
SD 6.15
Alpha .01H0: Mu = 18.98 vs the H1: Mu < 18.98A one tailed test
Degree of freedom = 25 – 1 = 24
Critical value t value = -2.492
Test statistics (t test) = (18.10-18.98)/(6.15/sqrt25) = -0.715
The management have a claim.
From the calculated t value, -0.715 is