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How to understand if $a-c lt b lt a+c$ then $b-c lt a lt b+c$
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I'm learning the concept of confidence interval. One of the important inference is that "the probability that sample mean is within two standard deviation of population mean equals to 95%." is equivalent to "there is a 95% probability that population mean is within two standard deviation of sample mean."
It seems the logic is if $a-c lt b lt a+c$ then to $b-c lt a lt b+c$ where a = population mean, b = sample mean and c = 2SD.
I don't understand.
algebra-precalculus statistics inequality confidence-interval
asked Aug 4 at 3:28
Kevin
133
add a comment |Â
up vote
2
down vote
favorite
I'm learning the concept of confidence interval. One of the important inference is that "the probability that sample mean is within two standard deviation of population mean equals to 95%." is equivalent to "there is a 95% probability that population mean is within two standard deviation of sample mean."
It seems the logic is if $a-c lt b lt a+c$ then to $b-c lt a lt b+c$ where a = population mean, b = sample mean and c = 2SD.
I don't understand.
algebra-precalculus statistics inequality confidence-interval
asked Aug 4 at 3:28
Kevin
133
I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm learning the concept of confidence interval. One of the important inference is that "the probability that sample mean is within two standard deviation of population mean equals to 95%." is equivalent to "there is a 95% probability that population mean is within two standard deviation of sample mean."
It seems the logic is if $a-c lt b lt a+c$ then to $b-c lt a lt b+c$ where a = population mean, b = sample mean and c = 2SD.
I don't understand.
algebra-precalculus statistics inequality confidence-interval
asked Aug 4 at 3:28
Kevin
133
I'm learning the concept of confidence interval. One of the important inference is that "the probability that sample mean is within two standard deviation of population mean equals to 95%." is equivalent to "there is a 95% probability that population mean is within two standard deviation of sample mean."
It seems the logic is if $a-c lt b lt a+c$ then to $b-c lt a lt b+c$ where a = population mean, b = sample mean and c = 2SD.
I don't understand.
algebra-precalculus statistics inequality confidence-interval
asked Aug 4 at 3:28
Kevin
133
edited 2 days ago
asked Aug 4 at 3:28
Kevin
133
asked Aug 4 at 3:28
Kevin
133
asked Aug 4 at 3:28
Kevin
133
133
I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03
add a comment |Â
I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03
I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03
I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
0
down vote
accepted
Let $mu$ and $barx$ be the population and sample means, respectively. The confidence interval:
$$barx-ME<mu<barx+ME (1)$$
where $ME=z_alpha/2fracbarx-musigma/sqrtn$ is the margin of error. Also, $ME=z_alpha/2SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$:
$$-barx+ME>-mu>-barx-ME (2)$$
Add $mu+barx$ to $(2)$:
$$mu+ME>barx>mu -ME (3)$$
Hence, $(1),(2),(3)$ are equivalent.
answered 2 days ago
farruhota
13.4k2632
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
add a comment |Â
up vote
1
down vote
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)land (b<a+c)$", which in turn is equivalent to $(a<b+c)land (b-c<a)$ by carrying the appropriate term to the other sides.
answered Aug 4 at 3:41
Saucy O'Path
2,264217
add a comment |Â
up vote
1
down vote
One way to see it is that if $a-c<b<a+c$ then $-a+c>-b>-a-c$ (multiplying by $-1$ reverses all the inequalities) and then adding $a+b$ to all terms gives $b+c>a>b-c$, which is what you want.
answered 2 days ago
Especially Lime
19k22252
add a comment |Â
up vote
1
down vote
Both formulae are equivalent to $|a - b| < c$.
answered 2 days ago
Alexey
639622
add a comment |Â
up vote
0
down vote
$$|hat p - mean| < 2*(SD)$$
means "the distance from $hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$hat p - mean < 2*(SD)$ or $hat p - mean > -2 * (standard deviations)$
therefore
$hat p < mean + 2*(SD)$ or $hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
answered Aug 4 at 3:40
Hanry Hu
1012
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $mu$ and $barx$ be the population and sample means, respectively. The confidence interval:
$$barx-ME<mu<barx+ME (1)$$
where $ME=z_alpha/2fracbarx-musigma/sqrtn$ is the margin of error. Also, $ME=z_alpha/2SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$:
$$-barx+ME>-mu>-barx-ME (2)$$
Add $mu+barx$ to $(2)$:
$$mu+ME>barx>mu -ME (3)$$
Hence, $(1),(2),(3)$ are equivalent.
answered 2 days ago
farruhota
13.4k2632
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
add a comment |Â
up vote
0
down vote
accepted
Let $mu$ and $barx$ be the population and sample means, respectively. The confidence interval:
$$barx-ME<mu<barx+ME (1)$$
where $ME=z_alpha/2fracbarx-musigma/sqrtn$ is the margin of error. Also, $ME=z_alpha/2SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$:
$$-barx+ME>-mu>-barx-ME (2)$$
Add $mu+barx$ to $(2)$:
$$mu+ME>barx>mu -ME (3)$$
Hence, $(1),(2),(3)$ are equivalent.
answered 2 days ago
farruhota
13.4k2632
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $mu$ and $barx$ be the population and sample means, respectively. The confidence interval:
$$barx-ME<mu<barx+ME (1)$$
where $ME=z_alpha/2fracbarx-musigma/sqrtn$ is the margin of error. Also, $ME=z_alpha/2SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$:
$$-barx+ME>-mu>-barx-ME (2)$$
Add $mu+barx$ to $(2)$:
$$mu+ME>barx>mu -ME (3)$$
Hence, $(1),(2),(3)$ are equivalent.
answered 2 days ago
farruhota
13.4k2632
Let $mu$ and $barx$ be the population and sample means, respectively. The confidence interval:
$$barx-ME<mu<barx+ME (1)$$
where $ME=z_alpha/2fracbarx-musigma/sqrtn$ is the margin of error. Also, $ME=z_alpha/2SE$, where $SE$ is the standard error.
Multiply $(1)$ by $-1$:
$$-barx+ME>-mu>-barx-ME (2)$$
Add $mu+barx$ to $(2)$:
$$mu+ME>barx>mu -ME (3)$$
Hence, $(1),(2),(3)$ are equivalent.
answered 2 days ago
farruhota
13.4k2632
answered 2 days ago
farruhota
13.4k2632
answered 2 days ago
farruhota
13.4k2632
answered 2 days ago
farruhota
13.4k2632
13.4k2632
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
add a comment |Â
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
@Kevin, Yes, the assumption is called the Central Limit Theorem, by which the sample size and the sampling distribution requirements must be satisfied. In solving problems on hypothesis testing (which is equivalent to the confidence interval), when the population SD is not known, the sample SD and $t$ table are recommended to be used instead.
– farruhota
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
Sorry, I deleted my question by accident. Here it is: "The standard deviation of the sample is different from that of the population (at least in most of the cases). We use the sample standard deviation here. Is this mean that the confidence interval is still a huge assumption?" Thank you!
– Kevin
2 days ago
add a comment |Â
up vote
1
down vote
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)land (b<a+c)$", which in turn is equivalent to $(a<b+c)land (b-c<a)$ by carrying the appropriate term to the other sides.
answered Aug 4 at 3:41
Saucy O'Path
2,264217
add a comment |Â
up vote
1
down vote
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)land (b<a+c)$", which in turn is equivalent to $(a<b+c)land (b-c<a)$ by carrying the appropriate term to the other sides.
answered Aug 4 at 3:41
Saucy O'Path
2,264217
add a comment |Â
up vote
1
down vote
up vote
1
down vote
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)land (b<a+c)$", which in turn is equivalent to $(a<b+c)land (b-c<a)$ by carrying the appropriate term to the other sides.
answered Aug 4 at 3:41
Saucy O'Path
2,264217
"$a-c<b<a+c$" is a placeholder for "$(a-c<b)land (b<a+c)$", which in turn is equivalent to $(a<b+c)land (b-c<a)$ by carrying the appropriate term to the other sides.
answered Aug 4 at 3:41
Saucy O'Path
2,264217
answered Aug 4 at 3:41
Saucy O'Path
2,264217
answered Aug 4 at 3:41
Saucy O'Path
2,264217
answered Aug 4 at 3:41
Saucy O'Path
2,264217
2,264217
add a comment |Â
add a comment |Â
up vote
1
down vote
One way to see it is that if $a-c<b<a+c$ then $-a+c>-b>-a-c$ (multiplying by $-1$ reverses all the inequalities) and then adding $a+b$ to all terms gives $b+c>a>b-c$, which is what you want.
answered 2 days ago
Especially Lime
19k22252
add a comment |Â
up vote
1
down vote
One way to see it is that if $a-c<b<a+c$ then $-a+c>-b>-a-c$ (multiplying by $-1$ reverses all the inequalities) and then adding $a+b$ to all terms gives $b+c>a>b-c$, which is what you want.
answered 2 days ago
Especially Lime
19k22252
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One way to see it is that if $a-c<b<a+c$ then $-a+c>-b>-a-c$ (multiplying by $-1$ reverses all the inequalities) and then adding $a+b$ to all terms gives $b+c>a>b-c$, which is what you want.
answered 2 days ago
Especially Lime
19k22252
One way to see it is that if $a-c<b<a+c$ then $-a+c>-b>-a-c$ (multiplying by $-1$ reverses all the inequalities) and then adding $a+b$ to all terms gives $b+c>a>b-c$, which is what you want.
answered 2 days ago
Especially Lime
19k22252
answered 2 days ago
Especially Lime
19k22252
answered 2 days ago
Especially Lime
19k22252
answered 2 days ago
Especially Lime
19k22252
19k22252
add a comment |Â
add a comment |Â
up vote
1
down vote
Both formulae are equivalent to $|a - b| < c$.
answered 2 days ago
Alexey
639622
add a comment |Â
up vote
1
down vote
Both formulae are equivalent to $|a - b| < c$.
answered 2 days ago
Alexey
639622
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Both formulae are equivalent to $|a - b| < c$.
answered 2 days ago
Alexey
639622
Both formulae are equivalent to $|a - b| < c$.
answered 2 days ago
Alexey
639622
answered 2 days ago
Alexey
639622
answered 2 days ago
Alexey
639622
answered 2 days ago
Alexey
639622
639622
add a comment |Â
add a comment |Â
up vote
0
down vote
$$|hat p - mean| < 2*(SD)$$
means "the distance from $hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$hat p - mean < 2*(SD)$ or $hat p - mean > -2 * (standard deviations)$
therefore
$hat p < mean + 2*(SD)$ or $hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
answered Aug 4 at 3:40
Hanry Hu
1012
add a comment |Â
up vote
0
down vote
$$|hat p - mean| < 2*(SD)$$
means "the distance from $hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$hat p - mean < 2*(SD)$ or $hat p - mean > -2 * (standard deviations)$
therefore
$hat p < mean + 2*(SD)$ or $hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
answered Aug 4 at 3:40
Hanry Hu
1012
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$|hat p - mean| < 2*(SD)$$
means "the distance from $hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$hat p - mean < 2*(SD)$ or $hat p - mean > -2 * (standard deviations)$
therefore
$hat p < mean + 2*(SD)$ or $hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
answered Aug 4 at 3:40
Hanry Hu
1012
$$|hat p - mean| < 2*(SD)$$
means "the distance from $hat p$ to the mean (which is symmetric, signless) is less than 2 standard deviations". Using inequality rules:
$hat p - mean < 2*(SD)$ or $hat p - mean > -2 * (standard deviations)$
therefore
$hat p < mean + 2*(SD)$ or $hat p > mean - 2 * (SD)$
or
$b < a + c$ or $b > a - c$
, as you would write it (your first expression).
You can also collect $a$ to one side and see how it affects $b$ and $c$.
The compound inequalities mean the same thing when you write with ors. But you write your equations as if these ors were ands, and that's why you're confused.
Great to learn about confidence intervals by the way.
answered Aug 4 at 3:40
Hanry Hu
1012
answered Aug 4 at 3:40
Hanry Hu
1012
answered Aug 4 at 3:40
Hanry Hu
1012
answered Aug 4 at 3:40
Hanry Hu
1012
1012
add a comment |Â
add a comment |Â
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I reformatted your post, could you please verify whether the inequalities within the title (or the body) were as you intended them to be? Further, what do you mean by "then to" in the title and the body?
– BLAZE
Aug 4 at 4:03