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Suppose that the purity (%) of a chemical solution y is related to the amount of a catalyst x (mg) by the linear regression model\ny=123-2.3x+\\epsi lon\nwhere \\epsi lon∼N(0,16).\na) What is the mean purity when the catalyst amount is 15mg ?\nb) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase?\nc) For a solution with catalyst amount x=20mg, find the probability that the purity is between 70 and 80 units. 1. Suppose that the purity%of a chemical solution y is related to the amount of a catalyst (mg by the linear regression model =1232.3x+c where~N(0,16) a) What is the mean purity when the catalyst amount is 15 mg? b) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase? c For a solution with catalyst amount =20 mg,find the probability that the purity is between 70 and 80 units. 

          Suppose that the purity (%) of a chemical solution y is related to the amount of a catalyst x (mg) by the linear regression model\ny=123-2.3x+\\epsi lon\nwhere \\epsi lon∼N(0,16).\na) What is the mean purity when the catalyst amount is 15mg ?\nb) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase?\nc) For a solution with catalyst amount x=20mg, find the probability that the purity is between 70 and 80 units.
1. Suppose that the purity%of a chemical solution y is related to the amount of a catalyst (mg by the linear regression model =1232.3x+c where~N(0,16)
a) What is the mean purity when the catalyst amount is 15 mg? b) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase? c For a solution with catalyst amount =20 mg,find the probability that the purity is between 70 and 80 units.
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suppose that the purity of a chemical solution y is related to the amount of a catalyst x mg by the linear regression modelny123 23xepsi lonnwhere epsi lonn016na what is the mean purity when 88356

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
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Suppose that the purity (%) of a chemical solution y is related to the amount of a catalyst x (mg) by the linear regression model\ny=123-2.3x+\\epsi lon\nwhere \\epsi lon∼N(0,16).\na) What is the mean purity when the catalyst amount is 15mg ?\nb) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase?\nc) For a solution with catalyst amount x=20mg, find the probability that the purity is between 70 and 80 units. 1. Suppose that the purity%of a chemical solution y is related to the amount of a catalyst (mg by the linear regression model =1232.3x+c where~N(0,16) a) What is the mean purity when the catalyst amount is 15 mg? b) What is the average change in purity for a 1-mg increase in catalyst? For a 10-mg increase? c For a solution with catalyst amount =20 mg,find the probability that the purity is between 70 and 80 units.
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Transcript

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00:01 Hi guys in this problem we need to find sample means x1 par and x2 par so as we know x1 par it's just the summation over x1 for x1 over n which is 10 so it's 57 .9 plus 62 plus 66 .2 and so on up to 71 okay over 10 so the sample mean is 65 .22.
00:35 Okay then x2 bar it's summation over x2 for x2 okay it's x1 not x i so over n which is also 10.
00:51 Okay so it's 66 .4 and so on up to 68 .8 over 10.
01:01 So this is 68 .42.
01:07 Okay.
01:09 Now to construct a 95 % confidence interval.
01:15 So we need to compute x1 bar minus x2 par, okay, plus or minus z of alpha over two times sigma 1 squared over n1 plus sigma 2 squared over n1 plus sigma 2 squared over n2 okay so after substitution we get this mu 1 minus mu 2 is more than or equal negative 5 .83 and less than or equal negative 0 .57 okay so now let's find z node so z node equal x1 part minus x2 minus delta node over the square root of sigma 1 squared over n1 plus sigma 2 squared over n okay so after substitution we can get this is negative 2 .39 okay so let's see the p value so the p value it's 2 times 1 minus 5 z node which is 2 .2.
02:42 Point 39 okay so it's 0 .0168 okay it's a zero value of mu 1 minus mu 2 is not in the this interval okay so it suggests that the mean active concentration depend on the choice of the catalyst okay for part c we need to find beta so beta is phi of z of alpha over 2 minus delta minus delta node over the square root of sigma 1 squared over n1 plus sigma 2 squared over n okay this is minus alpha of negative 1 .96 sorry i'm sorry so sorry minus minus delta minus delta node over the square root of sigma 1 squared over n 1 plus sigma 2 squared over n okay okay so this is 0 .03484 minus duo so it's 0 .03484 okay okay now let's find the power of the test...
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