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<v Instructor>Hello students,</v>

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welcome to Biostatistics ER, Example 2 Chapter 7.

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In this example,

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we will learn how to perform hypothesis testing

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for two sample matched continuous outcome data.

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The problem is from our textbook, Problem 18.

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The problem states,

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"Suppose the results of the analysis

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in problem 15 through Problem 17

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are reported and criticized because the participants

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were not randomized to different diets,

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and that there may be other factors

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associated with changes in cholesterol.

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A third study is run to estimate the effect

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of the low-carbohydrate diet on cholesterol levels.

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In the third study,

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participants' cholesterol levels

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are measured before starting the program,

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and then again after 6 months on the program.

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The data are shown here.

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Is there a significant increase in cholesterol

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after 6 months on the low-carbohydrate diet?

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Run the appropriate test at a 5% level of significance."

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So here, the samples are matched

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because the cholesterol levels of the same participants

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were measured before and after 6 months on the program.

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As we know, cholesterol level is also measured

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on a continuous scale,

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and here we will use t-test

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because our sample size is less than 30.

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With that, we will move to Step 1,

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which is set up hypothesis

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and determine level of significance.

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The null hypothesis here is mu d equal to 0

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and the alternative hypothesis is mu d greater than 0,

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and the level of significance is 0.05

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as stated in the problem.

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Here, I have provided you

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with the appropriate test statistics,

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which is a t-test.

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In Step 3, we will set up the decision rule

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and we will reject.

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Now if our t calculated is greater than or equal to 2.015,

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and this value was from the t-table

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based on degrees of freedom being 5.

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Now we are going to calculate the test statistics.

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So here, I strongly encourage you

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to either use Excel or SPSS

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to calculate the average

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and the standard deviation,

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and I have used Excel to calculate the average

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and the standard deviation.

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So once those values are calculated,

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we have to insert them into our formula.

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and the numerator here is going to be 5,

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which is the average,

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and the denominator is going to be the standard deviation,

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11.8 divided by the square root of sample size,

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which is 6.

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And once we perform the algebra,

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our t calculated is 1.04.

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So now in Step 5,

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and we are going to reach our conclusion.

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And here we will fail to reject null

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because 1.04 is less than 2.015.

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So we do not have statistically significant evidence

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at the alpha of 0.05 to show

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that there is a significant difference in cholesterol

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after 6 months on the diet.

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Thank you for your time and attention,

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and I'll see you in the next video.