How to calculate sample mean and variance given the confidence interval for the normal?
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A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).
What are the values of the sample mean and the variance? (X bar and S)
I got X bar to be 47.3.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
statistics normal-distribution standard-deviation variance confidence-interval
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up vote
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down vote
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A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).
What are the values of the sample mean and the variance? (X bar and S)
I got X bar to be 47.3.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
statistics normal-distribution standard-deviation variance confidence-interval
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).
What are the values of the sample mean and the variance? (X bar and S)
I got X bar to be 47.3.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
statistics normal-distribution standard-deviation variance confidence-interval
A random sample of size n=16 is taken from a random variable X~N(mu, sigma), with variance unknown. The 95% confidence interval for mu (44.7, 49.9).
What are the values of the sample mean and the variance? (X bar and S)
I got X bar to be 47.3.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
statistics normal-distribution standard-deviation variance confidence-interval
statistics normal-distribution standard-deviation variance confidence-interval
edited Nov 19 at 1:49
asked Nov 19 at 1:15
kmediate
115
115
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Yes, you have gotten the mean right, it is the average of the end points.
$$bar{x} - k S = 44.7$$
$$bar{x} + k S = 44.7$$
where $k$ is known to you.
Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, you have gotten the mean right, it is the average of the end points.
$$bar{x} - k S = 44.7$$
$$bar{x} + k S = 44.7$$
where $k$ is known to you.
Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
add a comment |
up vote
0
down vote
Yes, you have gotten the mean right, it is the average of the end points.
$$bar{x} - k S = 44.7$$
$$bar{x} + k S = 44.7$$
where $k$ is known to you.
Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, you have gotten the mean right, it is the average of the end points.
$$bar{x} - k S = 44.7$$
$$bar{x} + k S = 44.7$$
where $k$ is known to you.
Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.
Yes, you have gotten the mean right, it is the average of the end points.
$$bar{x} - k S = 44.7$$
$$bar{x} + k S = 44.7$$
where $k$ is known to you.
Since you already know $bar{x}$, you can now solve for $sigma$ in terms of $k$ and $bar{x}$.
edited Nov 19 at 1:38
answered Nov 19 at 1:20
Siong Thye Goh
95.7k1462116
95.7k1462116
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
add a comment |
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
I then got: 44.7 = 47.3 - S/(sqrt 16)* t(15,.025) for S=20.4
– kmediate
Nov 19 at 1:37
add a comment |
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