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Calculating a 95% confidence interval for the difference of two random variables
$begingroup$
Let $ x_1, ..., x_9 $ and $ y_1, ..., y_8 $ be two random samples of two populations.
$ bar x = 7 $ is the mean of the first and $ bar y = 11 $ the mean of the second sample. The sample standard deviations are $ s_x = 2 $ and $ s_y = 3.5 $. Now I want to calculate a 95% confidence interval for the difference of the mean of the two populations.
I know how to calculate a 95% confidence interval for the mean for the populations: $$ bar x -2*left(frac{s_x}{sqrt 9}right) $$ and $$ bar y -2*left(frac{s_y}{sqrt 8}right) $$ But don't know how to proceed from here.
probability statistics confidence-interval
asked Jan 29 at 13:49
Luca9984Luca9984
31
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add a comment |
$begingroup$
Let $ x_1, ..., x_9 $ and $ y_1, ..., y_8 $ be two random samples of two populations.
$ bar x = 7 $ is the mean of the first and $ bar y = 11 $ the mean of the second sample. The sample standard deviations are $ s_x = 2 $ and $ s_y = 3.5 $. Now I want to calculate a 95% confidence interval for the difference of the mean of the two populations.
I know how to calculate a 95% confidence interval for the mean for the populations: $$ bar x -2*left(frac{s_x}{sqrt 9}right) $$ and $$ bar y -2*left(frac{s_y}{sqrt 8}right) $$ But don't know how to proceed from here.
probability statistics confidence-interval
asked Jan 29 at 13:49
Luca9984Luca9984
31
$endgroup$
add a comment |
$begingroup$
Let $ x_1, ..., x_9 $ and $ y_1, ..., y_8 $ be two random samples of two populations.
$ bar x = 7 $ is the mean of the first and $ bar y = 11 $ the mean of the second sample. The sample standard deviations are $ s_x = 2 $ and $ s_y = 3.5 $. Now I want to calculate a 95% confidence interval for the difference of the mean of the two populations.
I know how to calculate a 95% confidence interval for the mean for the populations: $$ bar x -2*left(frac{s_x}{sqrt 9}right) $$ and $$ bar y -2*left(frac{s_y}{sqrt 8}right) $$ But don't know how to proceed from here.
probability statistics confidence-interval
asked Jan 29 at 13:49
Luca9984Luca9984
31
$endgroup$
Let $ x_1, ..., x_9 $ and $ y_1, ..., y_8 $ be two random samples of two populations.
$ bar x = 7 $ is the mean of the first and $ bar y = 11 $ the mean of the second sample. The sample standard deviations are $ s_x = 2 $ and $ s_y = 3.5 $. Now I want to calculate a 95% confidence interval for the difference of the mean of the two populations.
I know how to calculate a 95% confidence interval for the mean for the populations: $$ bar x -2*left(frac{s_x}{sqrt 9}right) $$ and $$ bar y -2*left(frac{s_y}{sqrt 8}right) $$ But don't know how to proceed from here.
probability statistics confidence-interval
probability statistics confidence-interval
asked Jan 29 at 13:49
Luca9984Luca9984
31
asked Jan 29 at 13:49
Luca9984Luca9984
31
asked Jan 29 at 13:49
Luca9984Luca9984
31
asked Jan 29 at 13:49
Luca9984Luca9984
31
asked Jan 29 at 13:49
Luca9984Luca9984
31
31
add a comment |
add a comment |
1 Answer
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$begingroup$
HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $bar{X}-bar{Y}$ is approximately normally distributed, with a sample mean of $bar{x}-bar{y}=7-11=-4$ and a standard deviation approximately equal to
$$s_{bar{x}-bar{y}}approx sqrt{frac{s_x^2}{n_x}+frac{s_y^2}{n_y}}$$
Can you use this to construct a confidence interval?
My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $bar{X}-bar{Y}$ is approximately normally distributed, with a sample mean of $bar{x}-bar{y}=7-11=-4$ and a standard deviation approximately equal to
$$s_{bar{x}-bar{y}}approx sqrt{frac{s_x^2}{n_x}+frac{s_y^2}{n_y}}$$
Can you use this to construct a confidence interval?
My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
add a comment |
$begingroup$
HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $bar{X}-bar{Y}$ is approximately normally distributed, with a sample mean of $bar{x}-bar{y}=7-11=-4$ and a standard deviation approximately equal to
$$s_{bar{x}-bar{y}}approx sqrt{frac{s_x^2}{n_x}+frac{s_y^2}{n_y}}$$
Can you use this to construct a confidence interval?
My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
add a comment |
$begingroup$
HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $bar{X}-bar{Y}$ is approximately normally distributed, with a sample mean of $bar{x}-bar{y}=7-11=-4$ and a standard deviation approximately equal to
$$s_{bar{x}-bar{y}}approx sqrt{frac{s_x^2}{n_x}+frac{s_y^2}{n_y}}$$
Can you use this to construct a confidence interval?
My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
$endgroup$
HINT: You have shown that you know how to find a confidence interval for the mean of a population, but not for the difference of the means of two populations. However, if $X$ and $Y$ are two random variables and a large number of samples is taken, then the random variable $bar{X}-bar{Y}$ is approximately normally distributed, with a sample mean of $bar{x}-bar{y}=7-11=-4$ and a standard deviation approximately equal to
$$s_{bar{x}-bar{y}}approx sqrt{frac{s_x^2}{n_x}+frac{s_y^2}{n_y}}$$
Can you use this to construct a confidence interval?
My only concern is that your sample sizes (of 8 and 9) are relatively small, so you may need to use the student’s t-distribution rather than the normal distribution.
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
answered Jan 29 at 13:55
FrpzzdFrpzzd
23k841110
23k841110
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
add a comment |
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
How did you come up with the formular for $ s_{bar{x}-bar{y}} $?
$endgroup$
– Luca9984
Jan 29 at 14:16
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
@Luca9984 It is just an approximation; like I said, you may have to use the t-distribution instead, since the approximation might not work as well with a small sample size.
$endgroup$
– Frpzzd
Jan 29 at 14:22
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
The formula to calculate a 95% confidence interval for a random sample contains the size of the random sample. What would be the size n in this case? 2?
$endgroup$
– Luca9984
Jan 29 at 14:33
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
$begingroup$
$n_x$ is the size of the sample from $X$, and $n_y$ is the size of the sample from $Y$.
$endgroup$
– Frpzzd
Jan 29 at 14:34
add a comment |
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