in this video we're going to talk about how to solve problems using the student's t-distribution table so let's focus on this one number one the average weight of 20 students in a certain school was found to be 165 pounds with a standard deviation of 4. 5 part a construct a 95 confidence interval for the population mean so let's write down what we know so we know that the average weight or the sample mean is 165 pounds the sample size is 20. the average weight of 20 students was measured in this study now the standard deviation is 4.
5 but here's a question for you what type of standard deviation are we dealing with is this the sample standard deviation or is this the population standard deviation what would you say now typical school is going to have more than 20 students a typical school might have a thousand students or 4 000 students so the 20 student in our study represents a small sample of all of the students in the school so therefore this is the sample standard deviation which is 4. 5 now when you have a question like this on a test you need to determine if you need to use a normal distribution or a student's t distribution to solve it now based on the title of this video you know it's going to be the the student's t distribution but you need to know why you should use which one when you know the population standard deviation typically you want to use the normal distribution but we don't have the population standard deviation we have the sample standard deviation and that's one indication that we need to use the student's t distribution to construct the confidence interval but there's something else that we need to consider as well and that is the sample size when n is equal to or greater than 30 then it's a good indication that you want to use the normal distribution but in this example n is less than 30. so that tells us that we want to use the student's t distribution instead so those are two good indicators that you should use the student's t distribution if you have the sample standard deviation as opposed to the population standard deviation and if n is less than 30.
now the population mean is going to be within the sample mean x bar plus or minus the t score this is t sub nu sub alpha over 2 which we'll talk about what that is times the sample standard deviation divided by the square root of n so we have x bar we know what s is and we know what n is in order to construct the confidence interval we need to determine the value of our t score or t value the subscript nu that's the greek letter nu looks like a v that's equal to n minus 1 which represents the degrees of freedom n is 20 in this example so 20 minus 1 is 19. so i'm going to write this here so we have 19 degrees of freedom alpha over 2 this is the one tell alpha the two tail alpha is just alpha by the way but our one tell alpha value alpha over two that's going to be one minus the confidence level divided by two now the confidence level is ninety-five percent but as a decimal if you divide it by 100 it's going to be 0. 95 1 minus 0.
95 is 0. 05 and if we divide that by 2 that's going to be 0. 025 so that is our one tau alpha value now just so you can get a visual illustration of this let's say we have a graph that looks like that i'm going to shade two regions so this area to the right this is alpha over two so that's our one tell alpha on the left this is also alpha over two combined if you add them together that would be alpha and in the middle this is one minus alpha so just know that alpha in this example is point zero five but alpha over two the one tell alpha is .
025 so now we could determine our t value so we have 19 degrees of freedom with a one tail alpha value of . 025 so at this point you want to go to the student's t distribution table and find the value that corresponds to a new value of 19 and an alpha over 2 value of 0. 025 and i'm going to show you how to do that briefly so here we have the student's t distribution table and we said that we had 19 degrees of freedom and an alpha over 2 value of 0.
025 so if we find the intersect in row and column it gives us this number 2. 093 and so that's going to be our t value when we have 19 degrees of freedom and an alpha over 2 value of 0. 025 it's equal to 2.
093 so now let's write our t value of 2. 093 here so now let's follow this expression to construct the confidence interval so let's replace x-bar with 165 our t value is 2. 093 and s is 4.
5 n is 20. 2. 093 times 4.
5 that's 9. 4185 and then divide that by the square root of 20 you're going to get 165 plus or minus 2. 106 now this number here has significance that is our ebm for the population mean which is also the margin of error so if you want to write this down the ebm for this type of problem is going to be our t value times s over the square root of n which gave us that number now to construct the 95 confidence interval it's going to be the sample mean minus the ebm value and then the sample mean plus the ebm value so once you get to this part you already have the answer for part b which is 2.
106 so now 165 minus 2. 106 that is 162. 8 and then 165 plus 2.
106 and that's 167. 106 so this right here is our 95 percent confidence interval so what does this mean this means that we're 95 confident that the average weight of all the students in this school is somewhere between 162. and 167.
106 so that's how you can construct a confidence interval using the student's t distribution table that's it for this problem number two a chemistry class at a certain university has 500 students the scores of 10 students were selected at random and are shown in the table below part a calculate the mean and standard deviation of the sample so let's start with that so the sample mean x bar is going to be the sum of all 10 scores of the students divided by the sample size so we're going to add 76 plus 84 plus 69 and so forth by the way if you want to try this problem feel free to pause the video and work on it and then when you're done you can just play the video to see if you have the same answer that i have so the sum that i got is 800 if we divided by 10 we get 80 which is a nice number to deal with so that's the sample mean that's the first part of part a now the next thing we need to do is calculate the standard deviation of the sample and so the formula that we're going to use to do that is going to be s is equal to the square root of the sum of x minus x bar squared over n minus 1. so x will be the individual values that we see here in the table x bar is 80. so the first value is 76 we're going to subtract it by the mean and then square the result the next value is 84 subtracted by the mean square the result the next value is 69 and then we're going to repeat the process until we include all 10 values so i'm going to put dot dot dot you could finish the rest and i'll put the last one as well so it's 77 minus 80 squared divided by n minus 1.
so that could take some time writing all of that out but if you want to show your work with less writing here's what you could do so 76 minus the mean 76 minus 80 is negative 4 but when you square negative 4 it's going to be positive 16 so i'm going to write 4 squared for the first one the second one 84 minus 80 that's going to be 4 and then we're going to square it as well now let's take this number and subtracted by the mean the difference between 69 and 80 is 11. 92 is 12 units from the mean 58 that's 22 units from the mean 89 is 9 units away from it so this is going to be 9 squared and then 73 is 7 units away from the mean once you square the negative signs won't matter 97 minus 80 that's 17 and then 85 minus 80 is 5. and then 77 minus 80 that's negative 3 but we're going to make it positive and then divide it by 9.
so go ahead and plug this in to your calculator and see what answer you get so i got for the sample standard deviation 11. 709 so that's the second part of part a so now let's move on to part b now let's also write that n is equal to 10. part b calculate the margin of error the ebm we could use this formula to calculate the ebm it's equal to our t value times this sample standard deviation over the square root of n so we need to calculate our t value first v that's n minus one so that's ten minus one which is nine so we have nine degrees of freedom and then we need to calculate our one tail alpha value which is one minus the confidence level divided by two so in part c we want to construct a ninety percent confidence interval that's going to be point nine one minus point nine is point 10 and point 10 divided by 2 is .
05 so our alpha over 2 value is going to be 0. 05 so now let's use that table that we used before and find the t value where we have nine degrees of freedom and a one tail alpha value of point zero five so here we're back to our students t distribution table and we have 9 degrees of freedom and an alpha over 2 value of 0. 05 so we can see that that's going to give us this number one point eight three three one so when we have nine degrees of freedom and an alpha over two value of point zero five this is going to be our t-score 1.
8331 so let's go ahead and write it here so now let's calculate the ebm value so it's 1. 8331 then times s which is 11. 709 divided by the square root of 10.
and so that comes out to be 6. 787 so that is the margin of error that's the error bound for the population mean that's it for part b now let's move on to part c construct a 90 confidence interval for the mean score of all the students in the chemistry class so we know the mean is going to be somewhere between x bar plus or minus our t value times s over the square root of n and so we know that x bar is 80 and all of this which is our ebm value that's 6. 787 so the 90 confidence interval is going to be between x bar minus ebm and the maximum value will be x bar plus the margin of error so this is going to be 80 minus 6.
787 and then 80 plus 6. 787 so 80 minus 6. 787 that's 73.
213 and then 80 plus that number that's going to be 86. 787 so this right here is our 90 confidence interval so we're 90 confident that the mean score of all the students in the chemistry class is somewhere between 73. 21 and 86.
