2 Continuing with Lawson Arena another important factor in m

2. Continuing with Lawson Arena, another important factor in maintaining good playing conditions is the humidity in the arena. Ideally the humidity should be at 30%. but anything less is also acceptable. Like the temperature, the university collected data on humidity. However. instead of once per day the humidity was taken twice per day. Over the two weeks the sample size of 28 had an average humidity of 32% and a standard deviation of 4%. The university wishes to test if the average humidity is less than 30%. Assume the humidity is normally distributed and that 95% confidence is needed. Treat % as a unit for this question. A. What is the null hypothesis? D. Calculate the proper test statistic. B. What is the alternative hypothesis? E. What is the statistical decision? C. What are the critical values? F. What is the answer to the actual question? 3. Using the information from problems 1 and 2 to answer the following: A. Determine the p-value associated with problem number 1. how does it compare to alpha? B. Calculate a 95% confidence interval on the average temperature from problem number 2. Would you reject the null? C. Determine the p-value associated with problem number 2. How does it compare to alpha? D. Calculate a 95% confidence interval on the average humidity from problem number 2. Would you reject the null? 4. Consider an experiment performed on a normally distributed population. A. Define what a type 1 error is. 13. Define what a type 2 error is. C. What is the power of the experiment if the probability of a type two error is 20%?

Solution

Q2.

Set Up Hypothesis
Null,University wishes to test the huidity is less than 0.30 H0: U<0.3
Alternate, University wishes to test the huidity is greater than 0.30 H1: U>0.3
Test Statistic
Population Mean(U)=0.3
Sample X(Mean)=0.32
Standard Deviation(S.D)=0.04
Number (n)=28
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =0.32-0.3/(0.04/Sqrt(27))
to =2.646
| to | =2.646
Critical Value
The Value of |t | with n-1 = 27 d.f is 1.703
We got |to| =2.646 & | t | =1.703
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.6458 ) = 0.00671
Hence Value of P0.05 > 0.00671,Here we Reject Ho

We decide that test the huidity is not less than 0.30