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    The question wants me to find the rang of values of \bar{x} for which the null hypothesis would be rejected.

    So far, I have:
    H_0:\mu = 100 

H_1:\mu \not= 100
    X - N(100, 0.36) (sorry couldn't figure out how to get ~. Latex is hard )

    To reject, P(Z<\frac{\bar{x}-100}{0.6}) \leq 0.025 or P(Z<\frac{\bar{x}-100}{0.6}) \geq 0.975

    I changed the first one to P(Z<\frac{100-\bar{x}}{0.6}) \leq 0.975, then used the tables.

    I end up with: \frac{100-\bar{x}}{0.6} \leq 1.960 or \frac{\bar{x}-100}{0.6} \geq 1.960

    Solving these give me \bar{x} \geq 98.82 and \bar{x} \geq 101.18. I'm guessing the first is meant to be \bar{x} \leq 98.82 and I've just messed up my inequalities somewhere but I can't quite see where. I think I could just change my inequalities to all equal signs which would just find the critical values and then use these to find the rejection region, but I'm just wondering where I went wrong.

    Thank you
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    (Original post by olivia30)
    ...
    It's \sim (\sim) to get \sim.

    Shouldn't the second inequality in your first line be P(Z > \alpha) \leq 0.025 or am I having a brain fart?
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    (Original post by Zacken)
    It's \sim (\sim) to get \sim.

    Shouldn't the second inequality in your first line be P(Z > \alpha) \leq 0.025 or am I having a brain fart?
    Thanks

    I think it's the same thing? Which should mean that I need to flip the inequality sign on the second line, which makes sense!
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    (Original post by olivia30)
    Thanks

    I think it's the same thing? Which should mean that I need to flip the inequality sign on the second line, which makes sense!
    For what it's worth, the way I would have done this is as follows:

    Let our test statistic be Z = \frac{\bar{X} - 100}{0.6}, then our critical values are given by: P(Z > k) = 0.025 whereupon a lookup in the percentage points table gives us Z \geq 1.96 and by symmetry Z \leq -1.96.
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    Maths.....

    Hmm... yes, I know some of these symbols..
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    (Original post by Zacken)
    For what it's worth, the way I would have done this is as follows:

    Let our test statistic be Z = \frac{\bar{X} - 100}{0.6}, then our critical values are given by: P(Z > k) = 0.025 whereupon a lookup in the percentage points table gives us Z \geq 1.96 and by symmetry Z \leq -1.96.
    Got it, thanks a ton

    To find the probability of a Type II error given \mu = 102, I want to do P(Z<\frac{101.18-102}{0.6}) - P(Z<\frac{98.82-102}{0.6}) right? It's just that this gives me 1 - P(Z<1.367) - 1 + P(Z<5.3) and our tables only go up to 3. Does this mean that P(Z<5.3) = 1? Sorry for all the questions
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    (Original post by olivia30)
    Got it, thanks a ton

    To find the probability of a Type II error given \mu = 102, I want to do P(Z<\frac{101.18-102}{0.6}) - P(Z<\frac{98.82-102}{0.6}) right? It's just that this gives me 1 - P(Z<1.367) - 1 + P(Z<5.3) and our tables only go up to 3. Does this mean that P(Z<5.3) = 1? Sorry for all the questions
    Don't fret it!

    Your tables only go up to 3 for a reason! It's because 3 as a z-score means 3 standard deviations away from the mean which is about as 100% as you get in the normal distribution.

    Now give me 5 minutes to go google up and learn about Type II errors before I answer your question because I have no clue what those are. :rofl:
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    (Original post by Zacken)
    Don't fret it!

    Your tables only go up to 3 for a reason! It's because 3 as a z-score means 3 standard deviations away from the mean which is about as 100% as you get in the normal distribution.

    Now give me 5 minutes to go google up and learn about Type II errors before I answer your question because I have no clue what those are. :rofl:
    You don't need to learn it just for me! It's just the probability of accepting the null hypothesis when it's actually false. I think that bit is right it's the P(Z<5.3) that confused me. Thank you so much!
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    (Original post by olivia30)
    You don't need to learn it just for me! It's just the probability of accepting the null hypothesis when it's actually false. I think that bit is right it's the P(Z<5.3) that confused me. Thank you so much!
    Okay, I've learnt about it. Want to check your answer with mine? :yep:

    By the way, I'm suspecting that maybe your 0.6 in the denominator might not be correct because of the weird numbers you're getting, do you mind giving me the information you used to calculate \frac{\sigma}{\sqrt{n}} so I can see if you did it right?
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    (Original post by Zacken)
    Okay, I've learnt about it. Want to check your answer with mine? :yep:

    By the way, I'm suspecting that maybe your 0.6 in the denominator might not be correct because of the weird numbers you're getting, do you mind giving me the information you used to calculate \frac{\sigma}{\sqrt{n}} so I can see if you did it right?
    Sure, I got 8.58%. Now watch as you get the correct answer and I don't, after I've been learning it for months and you've spent 10 mins on it

    It probably isn't then. The question says: "The resistances are normally distributed with mean \mu ohms and standard deviation 2.4 ohms. A random sample of 16 resistors is used to test the null hypothesis \mu = 100 "
    So I did X \sim N(100, \frac{2.4^2}{16}) which gives X \sim N(100,0.36) so rooting to get \sigma = 0.6.
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    (Original post by olivia30)
    Sure, I got 8.58%. Now watch as you get the correct answer and I don't, after I've been learning it for months and you've spent 10 mins on it

    It probably isn't then. The question says: "The resistances are normally distributed with mean \mu ohms and standard deviation 2.4 ohms. A random sample of 16 resistors is used to test the null hypothesis \mu = 100 "
    So I did X \sim N(100, \frac{2.4^2}{16}) which gives X \sim N(100,0.36) so rooting to get \sigma = 0.6.
    I get the same thing, yay! Guess OCR just like weird numbers.
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    (Original post by Zacken)
    I get the same thing, yay! Guess OCR just like weird numbers.
    Oh, great! :h: they do, they normally give really random numbers then every so often they will throw in a question right at the end of the paper that gives an integer value and I freak out thinking I've messed up because it's too nice :lol:

    Thank you for all your help!
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    (Original post by olivia30)
    Oh, great! :h: they do, they normally give really random numbers then every so often they will throw in a question right at the end of the paper that gives an integer value and I freak out thinking I've messed up because it's too nice :lol:

    Thank you for all your help!
    Darn, that sounds mean. :lol:

    No problem! My pleasure.
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    (Original post by Zacken)
    Darn, that sounds mean. :lol:

    No problem! My pleasure.
    If you get chance, could you possibly help me with the last bit of this question? All three parts have been a disaster :laugh:
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    (Original post by olivia30)
    If you get chance, could you possibly help me with the last bit of this question? All three parts have been a disaster :laugh:
    You're in luck, just got back home. :laugh: Shoot.
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    (Original post by Zacken)
    You're in luck, just got back home. :laugh: Shoot.
    The question is:
    "The company uses four different machines to manufacture the resistors. Three are correctly adjusted, so that they produce resistors with mean 100 ohms, but one is incorrectly adjusted, so that it produced resistors with mean 102 ohms. A sample of 16 resistors is selected at random from the output of one of the machines, chosen at random. Calculate the probability that the outcome of the test is to rejected the null hypothesis \mu = 100."

    I figured that I need to use 0.75 * something to do with the critical region in the first part + 0.25 * something to do with the second part, but I can't quite get there. Is the probability of rejecting if the mean is 100 just 5%? Then maybe for the one that is incorrectly adjusted, it's 1-0.0858? I don't think so :lol:

    Thank you!
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    (Original post by olivia30)

    I figured that I need to use 0.75 * something to do with the critical region in the first part + 0.25 * something to do with the second part, but I can't quite get there. Is the probability of rejecting if the mean is 100 just 5%? Then maybe for the one that is incorrectly adjusted, it's 1-0.0858? I don't think so :lol:

    Thank you!
    Good work! Impressive.

    To reject the null hypothesis on the correctly functioning machines, it would be a Type I error, which has a probability of 0.05.

    To reject the null hypothesis on the incorrectly functioning machine, it would be 1 minus the probability you worked out in the earlier part, as you rightly said.

    So you have 3/3 machines with a probability of 0.05 and 1/4 machines with a probability of approx 0.91 (no calc with me sorry) - what can you then do with this?
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    (Original post by Zacken)
    Good work! Impressive.

    To reject the null hypothesis on the correctly functioning machines, it would be a Type I error, which has a probability of 0.05.

    To reject the null hypothesis on the incorrectly functioning machine, it would be 1 minus the probability you worked out in the earlier part, as you rightly said.

    So you have 3/3 machines with a probability of 0.05 and 1/4 machines with a probability of approx 0.91 (no calc with me sorry) - what can you then do with this?
    So it'll be 0.75 * 0.05 + 0.25 * 1-0.0858 = 0.266 as you're choosing either a functioning one or a non-functioning one?
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    (Original post by olivia30)
    So it'll be 0.75 * 0.05 + 0.25 * 1-0.0858 = 0.266 as you're choosing either a functioning one or a non-functioning one?
    Yep. :yep:
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    (Original post by Zacken)
    Yep. :yep:
    Thank you so much!
 
 
 
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