7 7818 Interval estimation and hypothesis testing Set revised Nov 29 2010 You might want to read some of the chapter in MGB on Parametric Interval Estimation There are subtle di erences across questions Make sure to look for and understand those di erences 1 Assume the random variable X is normally distributed with unknown mean and variance 16 that is fX x 16 x Let X1 X2 Xn be a random sample from this population Determine Pr X 3 X 3 answer we know that Z Xp n is normally distributed with mean zero and variance of one so has no parameters In this case Z in this case Pr X 3 X 3 Pr X Pr Pr 3 0 X X p 4 n Xp 1 4 n So 3 3 X 3 p 0 p p 4 n 4 n 4 n 3 p 4 n X 3 p p 4 n 4 n X 3 3 p p p 4 n 4 n 4 n 3 3 p Z p Pr 4 n 4 n Pr This probability depends on the sample size but not on For any sample size we could look up the answer in the standard normal table For example if n 9 Pr 4 3pn Z 4 3pn Pr 4 3p9 Z 4 3p9 Pr 2 25 Z 2 25 NormalDist 2 25 NormalDist 2 25 0 975 55 What does this mean 97 555 of the random intervals X 3 to X 3 will contain giving us con dence that lies in this interval Note that this interval varies with X so is random If n 100 Pr 4 3pn Z 4 3pn Pr 4 p3100 Z 4 p3100 Pr 7 5 Z 7 5 NormalDist 7 5 NormalDist 7 5 1 000 0 and e ectively 100 of these random intervals will contain 1 While of knowing appears in Z Z is unit normal for all values of 1 Further note the sign cance 2 Assume the random variable X is normally distributed with unknown mean and variance 16 that is fX x 16 x Let X1 X2 Xn be a random sample from this population Determine Pr X 3 X 1 answer we know that Z Xp n is normally distributed with mean zero and variance of one and has no parameters In this case Z in this case Pr X 3 X Pr X Pr 1 3 0 X X p 4 n 3 X p p 4 n 4 n 1 3 Pr p Z p 4 n 4 n So 1 3 X p 0 p 4 n 4 n Pr Xp 4 n 1 p 4 n 1 p 4 n This probability depends on the sample size but not on For any sample size we could look up the answer in the standard normal table For example if n 9 Pr 4 1p9 Z 4 3p9 Pr 75 Z 2 25 NormalDist 2 25 NormalDist 75 0 214 4 What does this mean 21 44 of the random intervals X 3 to X 1 will contain In this problem we have derived a con dence interval on a parameter 3 Assume the random variable X is normally distributed with unknown mean and variance 16 that is fX x 16 x Let X1 X2 Xn be a random sample from this population Specify three di erent 4 con dence intervals for in terms of the sample average X answer we know that Z Xp n is normally distributed with mean zero and variance of one and has no parameters In this case Z Xp 4 n NormalDist Z NormalDist Z 4 Solution is f Z 0 524 4 g So one range on Z to get 40 of the distribution is 0 5244 to 0 5244 symmetrical around zero NormalDist Z NormalDist 0 4 Solution is f Z 1 281 6 g So another range on Z to get 40 of the distribution is 0 to 1 2816 NormalDist Z NormalDist 1 4 Solution is f Z 1 553 3 g So another range on Z to get 40 of the distribution is 0 1 to 1 5533 Note that these three ranges on Z are of increasing length and only the rst one is symmetric around zero Now we need to convert these ranges 2 in tems of Z into ranges in terms of X Start with the rst 4 Pr 5244 Z 5244 X p 5244 4 n p p 5244 4 n Pr 5244 4 n X p p Pr X 5244 4 n X 5244 4 n p p Pr X 5244 4 n X 5244 4 n p p Pr X 2 0976 n X 2 0976 n Pr 5244 p p So 40 of the random intervals X 2 0976 n to X 2 0976 nwill include Now consider the second 40 range on Z 4 Pr 0 Z 1 2816 Pr 0 X p 1 2816 4 n p Pr 0 X 1 2816 4 n p Pr X X 1 2816 4 n p Pr X 1 2816 4 n X p Pr X 5 1264 n X So 40 of the random intervals X consider the third 40 range on Z 4 p 5 1264 n to X will include Now Pr 1 Z 1 5533 X p 1 5533 4 n p p Pr 1 4 n X 1 5533 4 n p p X 1 5533 4 n Pr X 1 4 n p p Pr X 1 5533 4 n X 1 4 n p p Pr X 6 213 2 n X 4 n Pr 1 So 40 of the random intervals X p p 6 213 2 n to X 4 n will include Note that in this problem we have identi ed three di erent 4 con dence interval on the parameter u 4 Let G represent gubers where G is a random variable with fG g 125 if 0 g 4 25 if 4 g 6 0 otherwise 3 Charlotte tells you that she has randomly sampled one G from this distribution and its realized value is 7 How likely is it that she is either mistaken or lying Jesse tells you that he has randomly sampled one G from this distribution and its realized value is 5 6 How likely is it one will get a random draw from this distribution that is greater than or equal to 5 6 what is the probability of drawing a g 5 6 if the true distribution is fG g Explain answer The 7 could not have come from this population The 5 6 could have Picture fG g f g 0 25 0 2 0 15 0 1 0 05 0 0 2 4 6 8 g 50 of the density is in the range 4 g 6 Break this up into 5 equal ranges with 10 in each 4 g 4 4 4 4 g 4 8 4 8 g 5 2 5 2 g 5 6 and 5 6 g 6 0 So there is a 10 chance …