2. ERRORS: TYPE 1 (  ) TYPE 2 (  ) & POWER

3. Federal funds are to be allocated for financial aid to students for the cost of books. It is believe the mean cost of books to a college student is $300. We are concerned that it is really higher. So what do we do? We perform a hypothesis test. Ho: μ = 300 the mean cost of books is $300 Ha: μ > 300 the mean cost of books is > $300 

4. =325 Is the p-value <.01 ?   If the true mean is really higher then we will not be giving the students enough money to purchase books. So we take a random sample of 36 students who purchased books and the sample test data we collected has = 325  this Let  =.01

5. =325   p-value =.0186 Lets assume the null is correct µ = 300 and the sample test data we collected of = 325  gives a p-value =.0186 Let  =.01. So what do we do?

6. = 325   p-value =.0186 Since the p- value =.0186 not <   we do not reject the null.

7.   =325 p-value =.2023 If null is wrong and the true mean is really µ = $335 then the real p-value for = 325 is.2023.

8.   = p-value =  If null is wrong and the true mean is really µ = $335 then. So what is the probability to make a wrong decision? The red region below is known as β the probability of a type 2 error.

9.   H o :   p-value = ? p-value =.01 = ?  So by overlapping the two graphs we can see what to do. We need to find the data value that corresponds to the alpha=.01

10.   H o :   p-value = ? p-value =.01 = 328  The z score for alpha of.01 is 2.33 Solving this equation for we get a value of 328.

11.   H o :   p-value =.28 p-value =.01 = 328  We now need to find the p-value for 328 with the true μ = 335. This value is known as type 2 error. So the p-value for 328 with a true mean of 335 is.28 

12.   H o :   p-value =.28 α =.01 = 328  Below we see Alpha and Beta together.

13.   p-value =.28 p-value =.72 =328 POWER   H o :   Now when the null is wrong then the area to the right of 328 is the probability of making the correct decision. This is called power.

14.     p-value =.72 =328 POWER  We really do not know what the power is until you know the true value for 

15.   p-value =.97 =328 POWER  Now notice if  is even farther to the right of 300 then power gets even bigger.

16.     p-value =.97 = 328 POWER   

17. = 325   p-value =.0186 Now if .05 then we would have rejected a false null Since the p- value =.0186 <   reject the null. a =.05

18.   When   is small then  is big, if the null is false. So the probability of a type1 error is small, rejecting a truthful null. But the probability of making a type 2 error is big, failing to reject a false null.

19. When   is big then  is smaller. So the probability of a type1 error is bigger, rejecting a truthful null. But we reduce the probability of making a type 2 error. This area is smaller, failing to reject a false null.   Truth

20.  = ? Not knowing whether the null is true or false is what makes picking a correct alpha challenging. So you need to determine which type of error you do not want to make an then choose accordingly.

21.  = ? If you do not want to reject a truthful null pick a small   (Type 1 error) But you stand the chance of making a type two error. Failing to reject the null when it is false. 

22.  = ? If you want to reject a false null, which decreases the type 2 error pick a big   But you increase the chance of making a type one error. Rejecting the null when it is true. 

23.  = ? So looking at our problem what is the worse error? Type 1 reject a truthful null. So we would give the students more money when we did not need to. Big alpha will do this. Type 2 error. Failing to reject a false null. Not giving the students enough money. Small alpha will do this. So you need to determine which type of error you do not want to make an then choose accordingly.

24. A doctor thinks that a new diet will significantly increases the birth weight of babies. In 2002, the birth weight of full term babies were normally distributed with a mean of 7.53 pounds and a standard deviation of 1.15 pounds. The doctor randomly selects 50 recently pregnant mothers and persuades them to partake in the new diet. The mean weigh from these 50 babies 7.79 pounds. Is there sufficient evidence to support the claim the new diet will increase the birth weights of newborns? The significance level is 0.05. Write the null and alternative: Describe a type 1 and type 2 error. What is the probability for a type 1 error. We can not determine the probability of a type 2 error with out knowing the truth.

25. We are done

27. ERRORS: TYPE 1 (  ) TYPE 2 (  ) & POWER

28. 1.578  

29. 1.578  

30.   1.578

32.   1.578 Power

33.  Power 1.578

34.   Power

36. 1.578  p-value =.01

37.  1.578 p-value =.255 

38.  p-value =.255 p-value =.01 1.578 

39.  p-value =.255 p-value =.745 1.578 POWER  

40.  p-value =.745 1.578 POWER 

41.  p-value =.985 1.578 POWER 

42.  p-value =.985 1.578 POWER  