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<v Instructor>Hello, students.</v>

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Welcome to example five, chapter seven.

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In this example, we will learn how to perform

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hypothesis testing for two sample

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independent dichotomous outcome data.

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For this problem we will use problem 27 from our textbook.

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And the problem states, use the data shown in problem 25

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and test if there is a significant difference

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in the proportion of diabetic participants

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in the placebo group

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as compared to the standard drug group.

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Use a 5% level of significance.

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Now for your convenience, I have copied and pasted

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the table here and before we move forward to step one,

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I just want to clarify

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the two samples here are independent

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because we are comparing the placebo group

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to the standard drug group.

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And because we are evaluating proportion,

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the data here is dichotomous

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and given the sample size here, we will be using Z test.

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Now in step one, we are stating our hypotheses

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and we are also determining the level of significance.

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The null hypothesis here states that the proportion

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of diabetic participants is equal

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and the alternative hypothesis states that it is not equal.

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In step two, we will be selecting

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the appropriate test statistics

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and for step three we will be setting up our decision rule

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and we will reject the null hypothesis

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if our Z calculated is less than or equal to 1.960,

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or if our Z calculated is greater than

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or equal to 1.960.

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Now we will be computing the test statistics.

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Now here we will need to calculate

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the overall proportion p-hat

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and that will be x1 + x2 divided by n1 + n2.

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So we get those figures

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and we perform the calculation

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and our p-hat is 0.3.

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Now, when we substitute the values

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and we compute our test statistics,

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we see that our Z here is 0.49.

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Of course, the values here for p1-hat and p2-hat

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are provided to us in the table,

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and the p-hat is the one we just calculated

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and the n1 and the n2 are also provided to us.

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So again, once we insert these values into the formula

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and we perform the algebra,

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we find out that our Z calculated is equal to 0.49.

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Now as far as our conclusion is concerned,

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here, we will fail to reject the null

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because 0.49 is not less than

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or equal to negative 1.960

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and 0.49 is not greater than or equal to positive 1.960.

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Therefore, we do not have statistically significant evidence

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at alpha at 0.05 to show

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that there is a difference in the proportion

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of diabetic participants in the placebo group

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as compared to the standard drug group.

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Thank you for your time and attention

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and I'll see you in the next video.