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constructing a 95% confidence interval - manipulating inequalities
Clash Royale CLAN TAG#URR8PPP
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given the asymptotic distribution of $hattheta_1$ construct a 95% confidence interval for $theta$ for large samples:
$hattheta_1 = frachattheta_1-thetafracthetasqrtn$
I know that the confidence interval will be :
$-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$
and hence:
$frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$
Question:
How to get from $-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$ to $frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.
inequality estimation parameter-estimation confidence-interval
asked Jul 28 at 9:48
user1607
608
add a comment |Â
up vote
0
down vote
favorite
given the asymptotic distribution of $hattheta_1$ construct a 95% confidence interval for $theta$ for large samples:
$hattheta_1 = frachattheta_1-thetafracthetasqrtn$
I know that the confidence interval will be :
$-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$
and hence:
$frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$
Question:
How to get from $-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$ to $frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.
inequality estimation parameter-estimation confidence-interval
asked Jul 28 at 9:48
user1607
608
You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
given the asymptotic distribution of $hattheta_1$ construct a 95% confidence interval for $theta$ for large samples:
$hattheta_1 = frachattheta_1-thetafracthetasqrtn$
I know that the confidence interval will be :
$-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$
and hence:
$frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$
Question:
How to get from $-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$ to $frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.
inequality estimation parameter-estimation confidence-interval
asked Jul 28 at 9:48
user1607
608
given the asymptotic distribution of $hattheta_1$ construct a 95% confidence interval for $theta$ for large samples:
$hattheta_1 = frachattheta_1-thetafracthetasqrtn$
I know that the confidence interval will be :
$-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$
and hence:
$frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$
Question:
How to get from $-1.96 leq frachattheta_1-thetafracthetasqrtn leq 1.96$ to $frachattheta_11+frac1.96sqrtn leq theta leq frachattheta_11-frac1.96sqrtn$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.
inequality estimation parameter-estimation confidence-interval
asked Jul 28 at 9:48
user1607
608
asked Jul 28 at 9:48
user1607
608
asked Jul 28 at 9:48
user1607
608
asked Jul 28 at 9:48
user1607
608
608
You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57
add a comment |Â
You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57
You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57
You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57
add a comment |Â
2 Answers
2
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1
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accepted
For ease, let $hattheta_1=x, theta=y$.
Divide by $sqrtn$:
$$-frac1.96sqrtnle fracxy-1 le frac1.96sqrtn.$$
Add $1$:
$$1-frac1.96sqrtnle fracxy le 1+frac1.96sqrtn.$$
Raise to power $-1$:
$$frac11-frac1.96sqrtnge fracyx ge frac11+frac1.96sqrtn.$$
Note: $2<3 iff frac12>frac13$.
Now multiply by $x$ to get the final result.
answered Jul 28 at 14:45
farruhota
13.5k2632
add a comment |Â
up vote
0
down vote
Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $theta$ to one side and everything else to the other; divide by the coefficient of $theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $thetagt0$ and $nge4$, no reversal occurs. In the end, you can put the two inequalities back together.
answered Jul 28 at 10:01
joriki
164k10179328
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For ease, let $hattheta_1=x, theta=y$.
Divide by $sqrtn$:
$$-frac1.96sqrtnle fracxy-1 le frac1.96sqrtn.$$
Add $1$:
$$1-frac1.96sqrtnle fracxy le 1+frac1.96sqrtn.$$
Raise to power $-1$:
$$frac11-frac1.96sqrtnge fracyx ge frac11+frac1.96sqrtn.$$
Note: $2<3 iff frac12>frac13$.
Now multiply by $x$ to get the final result.
answered Jul 28 at 14:45
farruhota
13.5k2632
add a comment |Â
up vote
1
down vote
accepted
For ease, let $hattheta_1=x, theta=y$.
Divide by $sqrtn$:
$$-frac1.96sqrtnle fracxy-1 le frac1.96sqrtn.$$
Add $1$:
$$1-frac1.96sqrtnle fracxy le 1+frac1.96sqrtn.$$
Raise to power $-1$:
$$frac11-frac1.96sqrtnge fracyx ge frac11+frac1.96sqrtn.$$
Note: $2<3 iff frac12>frac13$.
Now multiply by $x$ to get the final result.
answered Jul 28 at 14:45
farruhota
13.5k2632
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For ease, let $hattheta_1=x, theta=y$.
Divide by $sqrtn$:
$$-frac1.96sqrtnle fracxy-1 le frac1.96sqrtn.$$
Add $1$:
$$1-frac1.96sqrtnle fracxy le 1+frac1.96sqrtn.$$
Raise to power $-1$:
$$frac11-frac1.96sqrtnge fracyx ge frac11+frac1.96sqrtn.$$
Note: $2<3 iff frac12>frac13$.
Now multiply by $x$ to get the final result.
answered Jul 28 at 14:45
farruhota
13.5k2632
For ease, let $hattheta_1=x, theta=y$.
Divide by $sqrtn$:
$$-frac1.96sqrtnle fracxy-1 le frac1.96sqrtn.$$
Add $1$:
$$1-frac1.96sqrtnle fracxy le 1+frac1.96sqrtn.$$
Raise to power $-1$:
$$frac11-frac1.96sqrtnge fracyx ge frac11+frac1.96sqrtn.$$
Note: $2<3 iff frac12>frac13$.
Now multiply by $x$ to get the final result.
answered Jul 28 at 14:45
farruhota
13.5k2632
answered Jul 28 at 14:45
farruhota
13.5k2632
answered Jul 28 at 14:45
farruhota
13.5k2632
answered Jul 28 at 14:45
farruhota
13.5k2632
13.5k2632
add a comment |Â
add a comment |Â
up vote
0
down vote
Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $theta$ to one side and everything else to the other; divide by the coefficient of $theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $thetagt0$ and $nge4$, no reversal occurs. In the end, you can put the two inequalities back together.
answered Jul 28 at 10:01
joriki
164k10179328
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
add a comment |Â
up vote
0
down vote
Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $theta$ to one side and everything else to the other; divide by the coefficient of $theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $thetagt0$ and $nge4$, no reversal occurs. In the end, you can put the two inequalities back together.
answered Jul 28 at 10:01
joriki
164k10179328
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $theta$ to one side and everything else to the other; divide by the coefficient of $theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $thetagt0$ and $nge4$, no reversal occurs. In the end, you can put the two inequalities back together.
answered Jul 28 at 10:01
joriki
164k10179328
Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $theta$ to one side and everything else to the other; divide by the coefficient of $theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $thetagt0$ and $nge4$, no reversal occurs. In the end, you can put the two inequalities back together.
answered Jul 28 at 10:01
joriki
164k10179328
edited Jul 28 at 11:58
answered Jul 28 at 10:01
joriki
164k10179328
answered Jul 28 at 10:01
joriki
164k10179328
answered Jul 28 at 10:01
joriki
164k10179328
164k10179328
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
add a comment |Â
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations
– user1607
Jul 28 at 11:19
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
@user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $Ale Ble C$ as two separate inequalities $Ale B$ and $Ble C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen?
– joriki
Jul 28 at 11:22
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
$$ -1.96 leq frachattheta_1-thetafracthetasqrtn$$ then $$ -1.96 fracthetasqrtn leq hattheta_1 - theta $$ and here i devide by -1.96 so $$ fracthetasqrtn + theta >= frachattheta_1-1.96$$
– user1607
Jul 28 at 11:32
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $theta$. What's the coefficient of $theta$ in $-1.96fracthetasqrt n+theta$?
– joriki
Jul 28 at 11:52
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
@user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $nge4$, since otherwise $1-1.96/sqrt n$ will be negative.
– joriki
Jul 28 at 11:59
add a comment |Â
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You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read.
– joriki
Jul 28 at 9:57