Students who took stat 1302 in 2023 can be broken into two...

Students who took Stat 1302 in 2023 can be broken into two groups; those who attended class regularly (population 1) and those who did not (population 2).  Suppose we wish to compare the average marks of these two groups on their first test. The standard deviation of the marks for both groups are known to be 12.98% and 12.47%.  A random sample of 20 marks from each group yielded means of 66.53% and 57.76% .  I suspect that the average for those who attended class on a regular basis may be significantly higher than those who do not. Let’s test my suspicion at the 10% level of significance

Transcript

00:01 So our question says a professor is concerned that two sections of college algebra that the teaches are not performing at the same level.

00:09 To test this claim, it looks at the main exam score for random sample of students from each of his classes.

00:15 In class one, the main exam score for 13 students is 81 .2, with a standard division of 7 .3.

00:22 And in class two the mean exam score for 18 students is 77 .3 with a standard deviation of 6 .1.

00:31 Test professor's claim at 0 .02 level of significant.

00:35 Assume that both populations are approximating normal and the population variances are equal.

00:41 Let class 1 be population 1 and let class 2 be population 2.

00:47 So let's move into our whiteboard.

00:50 So we have class 1 and we have class 2.

00:55 All right so for our class one the exam score for 13 students our n1 is a cost to 13 our x1 b is across to it's 1 .3 and the standard division s1 we have a value of 7 .3 all right this is it's 1 .2 so for our second class we have that in class 2 the main exam score for 18 students and that's n2 18 is 7 to 7 .3 so our n2 is 18 and x2 bar is equal to 77 .3 and the standard division of the second class we have a value of 6 .1 so the professor wants to test the claim that the algebra that it teaches the professor is concerned that the two sections of college algebra that it teaches are not performing at the same level so that means either one is doing better than the other or vice versa so let's state our null hypothesis so that is h1 is based on the fact that m u .1 is across to mule so this is all saying both classes are performing at the same level and the alternative hypothesis is based on the fact that mule 1 is actually not equals to mule so this is all saying both of them are not performing at the same level so after stating the null and alternative hypothesis the next step is for us to state the decision rule that will make us know whether we are supposed to accept or reject the null hypothesis and i'll be using the pay value method and it states that if our p value is greater than alpha we accept the null hypothesis and if our p values less than alpha we reject the null hypothesis where the level of significance is actually 0 .0 according to our question and that is 2 percent now the next thing is for us to get our test statistics and according to the question the data set is approximately normally distributed so it was mentioned that the data set is at least normally distributed.

03:06 So that means our test statistics is going to be a z test.

03:10 So we have our test statistics given towards a z is equal to x1 bar minus x2 bar divided by the square root of...

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