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ERRORS: TYPE 1 ( ) TYPE 2 ( ) & POWER

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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

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=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 Let =.01

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=325 p-value =.0225 Lets assume the null is correct µ = 300 and the sample test data we collected of = 325 s gives a p-value =.0225 Let =.01. So what do we do?

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= 325 p-value =.0225 Since the p- value =.0225 not < we do not reject the null. =.01 We can not say the mean cost of books is higher than $300

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p-value = ? p-value =.0225 325 But if the true cost is $335 then the distribution for it is below and the real p-value for 325 is a different value than.0186. The red region is the real p-value.

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=325 p-value =.2025 True mean But, if null is wrong and the true mean is µ = $335 then the real p-value for = 325 is.2023. P-value for = 335 is

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= p-value = If null is wrong and the true mean is really µ = $335 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 which depends on a

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H o : p-value = ? Sig. level =.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 the significance level.

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H o : p-value = ? Sig. level =.01 = 328 The z score for alpha of.01 is 2.33 Solving this equation for we get a value of 328. Recall our 325 is to the left of 328 in the red region.

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H o : p-value =.28 Sig. level =.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 True mean

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H o : p-value =.28 when = 335 α =.01 = 328 Below we see Alpha and Beta together.

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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. 1 - b

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p-value =.72 =328 POWER We really do not know what the power is because we do not know what is until you know the true value for

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p-value =.97 =328 POWER Now notice if the true is even farther to the right of 300 then power gets even bigger.

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p-value =.97 = 328 POWER

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= 325 p-value =.0225 Now if.05 then we would have rejected a false null Since the p- value =.0225 < reject the null. a =.05

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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.

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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

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= ? 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. Truth

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= ? 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.

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= ? 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.

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= ? 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.

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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.

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If the true weight was 8lbs, what is the probability of the type 2 error?

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Go to Notebook

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We are done

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ERRORS: TYPE 1 ( ) TYPE 2 ( ) & POWER

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1.578

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1.578

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1.578

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1.578 Power

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Power 1.578

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Power

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1.578 p-value =.01

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1.578 p-value =.255

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p-value =.255 p-value =.01 1.578

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p-value =.255 p-value =.745 1.578 POWER

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p-value =.745 1.578 POWER

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p-value =.985 1.578 POWER

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p-value =.985 1.578 POWER

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UNIT 6 CHAPTERS 23,24,25 By Courtney and Ayla. CHAPTER 23 In this chapter we discuss the true means of one scenario. We compare a given mean for a scenario.

UNIT 6 CHAPTERS 23,24,25 By Courtney and Ayla. CHAPTER 23 In this chapter we discuss the true means of one scenario. We compare a given mean for a scenario.

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