Stats 612 Lecture 3 Notes1

# Stats 612 Lecture 3 Notes1 - Lecture 3 First let us begin...

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Lecture 3, First let us begin with and example of a 2 tailed Confidence Interval X ( - ) xi x 2 40 100 50 0 60 100 45 25 55 25 50 0 300 250 = x 50 = n 6 =( - ) - = = s2 xi x 2n 1 2505 50 = = = . = . sx sn 506 8 67 2 94 - ∝ < < + ∝ = - Px t 2sx μ x t 2sx 1 α - . . < < + . ( . )= - P50 2 5712 94 μ 50 2 571 2 94 1 α . < < . = - P42 44 μ 57 56 1 α In the above problem we ended up adding and subtracting a value of 7.56 to the sample mean of 50 in order to get our confidence interval. Imagine that the boss says that value is too large. He/she wants a narrower confidence interval. The boss wants the value that you add and subtract to be 4. That is she wants a guarantee that the population mean is at least within 4 units of the sample mean, rather than 7.56. That value of four that I mentioned would be called the “margin of error.” 1

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The boss is specifying that the margin of error is to be no bigger than 4 with a probability of 0.95, for example. Also, in the following example, I have assumed that σ 2 is equal to 50. The square root of 50 is 7.07. Margin of Error = ( )= ; = ∝ If z 2σn E often called permissible error 4 then n z 2σE2 = [1.96*7.07/4] 2 = 601 That is to say, if we take a sample of size 601, we will be able to guarantee that the population mean is within four units of our sample mean with a probability of .95. Nature of the alternative Hypothesis, also called the research hypothesis If area of concern is to improve a product (make it last longer in days) , then the research hypothesis (the alternative hypothesis) becomes H a : µ >X, where X is specified to be some specific value and the null hypothesis would be H o : µ ≤ X. Example of testing a research hypothesis: H o : µ ≥ 8 H a : µ < 8 Imagine that we are testing a new cold drug. The manufactures are claiming that with this new drug, a patient will be relieved of the symptoms of shingles within less than eight days. That is the research hypothesis. We calculate P. If the value of P < α reject H 0 : This rule is always the same no matter the null or alternative hypothesis. A low p value indicates that the probability of this happening when the null hypothesis is true is very small. Yet we know that the sample finding did happen, even though according to the null hypothesis it is not supposed to happen. Therefore we throw out the null hypothesis. I hope that makes sense. The following is our sample findings. = x 7 2
σ = 2 n = 100 So: = =. σx 210 2 (recall = σx σn ) =( - ) =( - ). =- z x μ σx 7 8 2 5 Graph 1: The p value is extremely low. There is virtually no area to the left of -5 standard errors n =100 P value X = 7 8 -5 NB: the sign of the z value tells you what side of the hypothesized population mean, that the sample mean fell on. The p value associated with -5 is extremely close to zero, this value will not occur by chance alone. This means that which occurred is extremely rare (or very improbable). This value of p is certainly less than alpha. We no longer accept the null hypothesis. We accept the alternative.

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