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CaptainBlack said: I think that depends on what you gave been asked to do this for. If this is for a test to accept or reject the null hypothesis, then no. Click to expand... But how to find the power of the test? Give me a hint please
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Leona said: Can I please ask you a couple of more noob questions? I've found $$\displaystyle c =1.65 * sqrt(14) +7 =13.2$$, do I have to calculate it for H1 too? And what are the steps to find the power of the test for obtained c? Thank you in advance.. Click to expand... I think that depends on what you gave been asked to do this for. If this is for a test to accept or reject the null hypothesis, then no.
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Can I please ask you a couple of more noob questions? I've found $$\displaystyle c =1.65 * sqrt(14) +7 =13.2$$, do I have to calculate it for H1 too? And what are the steps to find the power of the test for obtained c? Thank you in advance..
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CaptainBlack said: Under the null hypothesis you have (approximatly) $x_n \sim N(7,14)$ You need the critical value $c$ such that $P(x_n>c)=0.05$. The z-score corresponding to this is $F(z)=0.95$ where $F(.)$ is the standard normal cumulative distribution function. This value can be looked up in a table or found using suitable software or just asking a search engine. The z-score corresponding to $c$ is $z=(c-7)/\sqrt{14}$, so given the z-score corresponding to the confidence level you solve this for $c$ Click to expand... Thank you so much!
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Under the null hypothesis you have (approximatly) $x_n \sim N(7,14)$ You need the critical value $c$ such that $P(x_n>c)=0.05$. The z-score corresponding to this is $F(z)=0.95$ where $F(.)$ is the standard normal cumulative distribution function. This value can be looked up in a table or found using suitable software or just asking a search engine. The z-score corresponding to $c$ is $z=(c-7)/\sqrt{14}$, so given the z-score corresponding to the confidence level you solve this for $c$
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CaptainBlack said: Lets hope this is right this time Under the null hypothesis $x_n$ is the sum of $7$ independent identically distributed RVs $Y=(X-5)^2$ with mean $$\mu_Y=\overline{(X-5)^2}=\sigma^2=1$$ and variance $$v_Y=E((Y-\mu_Y)^2)=E(Y^2-2\mu_Y Y+\mu_Y^2)=m_4(X)-2 \mu_Y^2+\mu_Y^2$$ where $m_4(X)$ is the fourth central moment of $X$ and as $X$ has a normal distribution is equal to $3\sigma^4$, so: $$v_Y=3-2+1=2$$ Now because $x_n$ is the sum of seven iid RVs with means $\mu_Y=1$ and variance $v_Y=2$ we know from the CLT that $x_n$ is approximatle $\sim N(7\mu_Y, 7v_Y)$ Numerical/Monte-Carlo check on this: Code: x=randn(1000000,7)+5;xx=sum((x'-5).^2)';mean(xx),var(xx) ans = 6.9945 ans = 14.004 Click to expand... Thank you! Could you please advise me some steps to solving it?
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SupremeCookie said: Notice: $$\sigma^2=\frac{\sum_{i=1}^7(x_i-\mu)^2}{n}=\frac{(x_1-5)^2 + (x_2-5)^2 +\dots+ (x_7-5)^2}{7}$$ Click to expand... Thank you! I'm really bad in this topic, what should I do next? Should I use t-test or z-test? Could you kindly advise me solution steps?
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Leona said: Hello everyone! Could you please advise me how to solve this? View attachment 44587 Click to expand... Lets hope this is right this time Under the null hypothesis $x_n$ is the sum of $7$ independent identically distributed RVs $Y=(X-5)^2$ with mean $$\mu_Y=\overline{(X-5)^2}=\sigma^2=1$$ and variance $$v_Y=E((Y-\mu_Y)^2)=E(Y^2-2\mu_Y Y+\mu_Y^2)=m_4(X)-2 \mu_Y^2+\mu_Y^2$$ where $m_4(X)$ is the fourth central moment of $X$ and as $X$ has a normal distribution is equal to $3\sigma^4$, so: $$v_Y=3-2+1=2$$ Now because $x_n$ is the sum of seven iid RVs with means $\mu_Y=1$ and variance $v_Y=2$ we know from the CLT that $x_n$ is approximatle $\sim N(7\mu_Y, 7v_Y)$ Numerical/Monte-Carlo check on this: Code: x=randn(1000000,7)+5;xx=sum((x'-5).^2)';mean(xx),var(xx) ans = 6.9945 ans = 14.004
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Notice: $$\sigma^2=\frac{\sum_{i=1}^7(x_i-\mu)^2}{n}=\frac{(x_1-5)^2 + (x_2-5)^2 +\dots+ (x_7-5)^2}{7}$$
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