Mixing
Use the Central Limit Theorem to determine the critical region corresponding to significance level
$begingroup$
Let $X_1;...;X_n$ be a random sample from a $U(0;theta)$ distribution with $theta>0$. Define the random variable $T = overline{X}_n$.
Let n= 144. You want to test $H_0$: $theta= 12$ against $H_1$: $theta neq 12$ and use $T$ as your test statistic. Suppose $overline{x}_{144} = 6.5$.
Use the Central Limit Theorem to determine the critical region corresponding to significance level $alpha = 0.05$ and report your conclusion about the null hypothesis.
In order to calculate the critical region (value of T where we reject $H_0$ in favour of $H_1$) I have to determine the critical values, but how can I do it? Can someone help me giving some hints, Thanks!
Update: (this is what I think I should do, where I'm wrong?)
$P(Tgeq C_1)=0.05$ From the $N(0,1)$ table $C=1.64$.
$P(Tleq C_1)=0.95$ From the $N(0,1)$ table $C=-1.64$.
$Z = frac{overline{X_n}-mu}{frac{sigma}{sqrt{n}}}$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=1.64:overline{X_n}=7.64:K=(7.64;infty)$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=-1.64:overline{X_n}=4.36:K=(infty;4.36)$
statistics
asked Jan 10 at 11:58
FTACFTAC
2719
$endgroup$
add a comment |
$begingroup$
Let $X_1;...;X_n$ be a random sample from a $U(0;theta)$ distribution with $theta>0$. Define the random variable $T = overline{X}_n$.
Let n= 144. You want to test $H_0$: $theta= 12$ against $H_1$: $theta neq 12$ and use $T$ as your test statistic. Suppose $overline{x}_{144} = 6.5$.
Use the Central Limit Theorem to determine the critical region corresponding to significance level $alpha = 0.05$ and report your conclusion about the null hypothesis.
In order to calculate the critical region (value of T where we reject $H_0$ in favour of $H_1$) I have to determine the critical values, but how can I do it? Can someone help me giving some hints, Thanks!
Update: (this is what I think I should do, where I'm wrong?)
$P(Tgeq C_1)=0.05$ From the $N(0,1)$ table $C=1.64$.
$P(Tleq C_1)=0.95$ From the $N(0,1)$ table $C=-1.64$.
$Z = frac{overline{X_n}-mu}{frac{sigma}{sqrt{n}}}$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=1.64:overline{X_n}=7.64:K=(7.64;infty)$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=-1.64:overline{X_n}=4.36:K=(infty;4.36)$
statistics
asked Jan 10 at 11:58
FTACFTAC
2719
$endgroup$
1
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29
add a comment |
$begingroup$
Let $X_1;...;X_n$ be a random sample from a $U(0;theta)$ distribution with $theta>0$. Define the random variable $T = overline{X}_n$.
Let n= 144. You want to test $H_0$: $theta= 12$ against $H_1$: $theta neq 12$ and use $T$ as your test statistic. Suppose $overline{x}_{144} = 6.5$.
Use the Central Limit Theorem to determine the critical region corresponding to significance level $alpha = 0.05$ and report your conclusion about the null hypothesis.
In order to calculate the critical region (value of T where we reject $H_0$ in favour of $H_1$) I have to determine the critical values, but how can I do it? Can someone help me giving some hints, Thanks!
Update: (this is what I think I should do, where I'm wrong?)
$P(Tgeq C_1)=0.05$ From the $N(0,1)$ table $C=1.64$.
$P(Tleq C_1)=0.95$ From the $N(0,1)$ table $C=-1.64$.
$Z = frac{overline{X_n}-mu}{frac{sigma}{sqrt{n}}}$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=1.64:overline{X_n}=7.64:K=(7.64;infty)$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=-1.64:overline{X_n}=4.36:K=(infty;4.36)$
statistics
asked Jan 10 at 11:58
FTACFTAC
2719
$endgroup$
Let $X_1;...;X_n$ be a random sample from a $U(0;theta)$ distribution with $theta>0$. Define the random variable $T = overline{X}_n$.
Let n= 144. You want to test $H_0$: $theta= 12$ against $H_1$: $theta neq 12$ and use $T$ as your test statistic. Suppose $overline{x}_{144} = 6.5$.
Use the Central Limit Theorem to determine the critical region corresponding to significance level $alpha = 0.05$ and report your conclusion about the null hypothesis.
In order to calculate the critical region (value of T where we reject $H_0$ in favour of $H_1$) I have to determine the critical values, but how can I do it? Can someone help me giving some hints, Thanks!
Update: (this is what I think I should do, where I'm wrong?)
$P(Tgeq C_1)=0.05$ From the $N(0,1)$ table $C=1.64$.
$P(Tleq C_1)=0.95$ From the $N(0,1)$ table $C=-1.64$.
$Z = frac{overline{X_n}-mu}{frac{sigma}{sqrt{n}}}$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=1.64:overline{X_n}=7.64:K=(7.64;infty)$
$frac{overline{X_n}-6}{frac{12}{sqrt{144}}}=-1.64:overline{X_n}=4.36:K=(infty;4.36)$
statistics
statistics
asked Jan 10 at 11:58
FTACFTAC
2719
asked Jan 10 at 11:58
FTACFTAC
2719
edited Jan 10 at 14:55
FTAC
asked Jan 10 at 11:58
FTACFTAC
2719
asked Jan 10 at 11:58
FTACFTAC
2719
asked Jan 10 at 11:58
FTACFTAC
2719
2719
1
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29
add a comment |
1
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29
1
1
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Some hints:
- According to CLT you have $Z = frac{overline{X_n}-mu_{theta}}{frac{sigma_{theta}}{sqrt{n}}} stackrel{approx}{sim} N(0,1)$.
$H_0: theta = 12 stackrel{X sim U(0,theta)}{Longrightarrow} mu_{theta}=frac{theta}{2} = 6$ and $sigma_{theta}^2 =frac{theta^2}{12}$.- As $H_1: theta neq 12$, you have a 2-tailed test and find the $z$-score by
$$P(z_{frac{alpha}{2}}leq Z leq z_{1-frac{alpha}{2}}) = 1- alpha = 0.95$$. - The critical region results by backwards calculation from $z < z_{frac{alpha}{2}}$ and $z > z_{1-frac{alpha}{2}}$.
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
$endgroup$
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
|
show 3 more comments
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$begingroup$
Some hints:
- According to CLT you have $Z = frac{overline{X_n}-mu_{theta}}{frac{sigma_{theta}}{sqrt{n}}} stackrel{approx}{sim} N(0,1)$.
$H_0: theta = 12 stackrel{X sim U(0,theta)}{Longrightarrow} mu_{theta}=frac{theta}{2} = 6$ and $sigma_{theta}^2 =frac{theta^2}{12}$.- As $H_1: theta neq 12$, you have a 2-tailed test and find the $z$-score by
$$P(z_{frac{alpha}{2}}leq Z leq z_{1-frac{alpha}{2}}) = 1- alpha = 0.95$$. - The critical region results by backwards calculation from $z < z_{frac{alpha}{2}}$ and $z > z_{1-frac{alpha}{2}}$.
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
$endgroup$
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
|
show 3 more comments
$begingroup$
Some hints:
- According to CLT you have $Z = frac{overline{X_n}-mu_{theta}}{frac{sigma_{theta}}{sqrt{n}}} stackrel{approx}{sim} N(0,1)$.
$H_0: theta = 12 stackrel{X sim U(0,theta)}{Longrightarrow} mu_{theta}=frac{theta}{2} = 6$ and $sigma_{theta}^2 =frac{theta^2}{12}$.- As $H_1: theta neq 12$, you have a 2-tailed test and find the $z$-score by
$$P(z_{frac{alpha}{2}}leq Z leq z_{1-frac{alpha}{2}}) = 1- alpha = 0.95$$. - The critical region results by backwards calculation from $z < z_{frac{alpha}{2}}$ and $z > z_{1-frac{alpha}{2}}$.
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
$endgroup$
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
|
show 3 more comments
$begingroup$
Some hints:
- According to CLT you have $Z = frac{overline{X_n}-mu_{theta}}{frac{sigma_{theta}}{sqrt{n}}} stackrel{approx}{sim} N(0,1)$.
$H_0: theta = 12 stackrel{X sim U(0,theta)}{Longrightarrow} mu_{theta}=frac{theta}{2} = 6$ and $sigma_{theta}^2 =frac{theta^2}{12}$.- As $H_1: theta neq 12$, you have a 2-tailed test and find the $z$-score by
$$P(z_{frac{alpha}{2}}leq Z leq z_{1-frac{alpha}{2}}) = 1- alpha = 0.95$$. - The critical region results by backwards calculation from $z < z_{frac{alpha}{2}}$ and $z > z_{1-frac{alpha}{2}}$.
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
$endgroup$
Some hints:
- According to CLT you have $Z = frac{overline{X_n}-mu_{theta}}{frac{sigma_{theta}}{sqrt{n}}} stackrel{approx}{sim} N(0,1)$.
$H_0: theta = 12 stackrel{X sim U(0,theta)}{Longrightarrow} mu_{theta}=frac{theta}{2} = 6$ and $sigma_{theta}^2 =frac{theta^2}{12}$.- As $H_1: theta neq 12$, you have a 2-tailed test and find the $z$-score by
$$P(z_{frac{alpha}{2}}leq Z leq z_{1-frac{alpha}{2}}) = 1- alpha = 0.95$$. - The critical region results by backwards calculation from $z < z_{frac{alpha}{2}}$ and $z > z_{1-frac{alpha}{2}}$.
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
edited Jan 10 at 15:06
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
answered Jan 10 at 12:19
trancelocationtrancelocation
10.8k1723
10.8k1723
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
|
show 3 more comments
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
1
1
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
$begingroup$
Because the sample mean may deviate too much to the left or to the right from the expected mean. So, there must be a lower critical value and an upper one. (2-tailed test).
$endgroup$
– trancelocation
Jan 10 at 14:40
1
1
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
$begingroup$
You forgot to take the square root while calculating $sigma$. It is $sqrt{12} = 2sqrt{3}$. I got the critical values as $x_L approx 5.4342$ and $x_R approx 6.5658$.
$endgroup$
– trancelocation
Jan 10 at 14:59
1
1
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
$begingroup$
Besides this, you need to distribute $alpha = 0.05$ equally on both sides. So, the $z$-scores are $z_{0.025} approx -1.96$ and $z_{0.975} approx 1.96$.
$endgroup$
– trancelocation
Jan 10 at 15:03
1
1
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
$begingroup$
I adjusted the indices of the $z$-scores for more clarity. Btw., I will be off for a while. Will check in here later.
$endgroup$
– trancelocation
Jan 10 at 15:08
1
1
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
$begingroup$
@FabioTaccaliti You are right concerning the usage of $z_{alpha}$ and $z_{frac{alpha}{2}}$ in 1- and 2-tailed tests. But the $p$ value itself is related to a specific sample and its $z$-score say $z_{test}$. It says how probable an even more deviating test result is provided that $H_0$ is true: In the 2-tailed case you have: $p = P(|Z| > z_{test}; | ; H_0)$. So, for example, the boundaries $x_L$ and $x_R$ of the critical region have a $p$-value of $0.05$.
$endgroup$
– trancelocation
Jan 11 at 5:06
|
show 3 more comments
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1
$begingroup$
Are you sure that $T = X_n$ and not $T = overline{X_n}$?
$endgroup$
– trancelocation
Jan 10 at 12:05
$begingroup$
Yes you are right, fixed!
$endgroup$
– FTAC
Jan 10 at 14:29