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    "The time it takes for a cheese and ham sandwhich to be toasted in a Cass ground floor cafetaria appears to be normally distributed. A random sample of the times for toasting 20 such sandwhich had a sample mean of 151 seconds and a sample standard deviation of 21.9 seconds.


    What is the value of the upper end point of the 99% confidence interval for the mean time to toast a cheese and ham sandwich"


    I got 163.61 but apparently the answer is 165.0098, any idea how they got this?.
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    (Original post by Mentally)
    "The time it takes for a cheese and ham sandwhich to be toasted in a Cass ground floor cafetaria appears to be normally distributed. A random sample of the times for toasting 20 such sandwhich had a sample mean of 151 seconds and a sample standard deviation of 21.9 seconds."

    I got 163.61 but apparently the answer is 165.0098, any idea how they got this?.
    Am I being thick or have you not given the question? What are we meant to be finding?
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    (Original post by Zacken)
    Am I being thick or have you not given the question? What are we meant to be finding?
    No i was being thick, sec lemme edit
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    (Original post by Mentally)
    "The time it takes for a cheese and ham sandwhich to be toasted in a Cass ground floor cafetaria appears to be normally distributed. A random sample of the times for toasting 20 such sandwhich had a sample mean of 151 seconds and a sample standard deviation of 21.9 seconds."

    I got 163.61 but apparently the answer is 165.0098, any idea how they got this?.
    What's the question.
    So  X \sim N(151, 21.9^2) .
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    (Original post by B_9710)
    What's the question.
    So  X~N(151, 21.9^2) .
    see op now
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    (Original post by Mentally)

    I got 163.61 but apparently the answer is 165.0098, any idea how they got this?.
    I got 163.61 too and I'm inclined to agree with you.
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    (Original post by Zacken)
    I got 163.61 too and I'm inclined to agree with you.
    OK how about this one

    "A sample of 24 independent observation is taken from a normal distribution of unknown mean μ but known variance σ2 = 96.42.
    The sample mean is 61.88 and sample variance is 57. What is the width of the 98% confidence interval for μ ?"
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    (Original post by Zacken)
    I got 163.61 too and I'm inclined to agree with you.
    Same.
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    (Original post by Mentally)
    OK how about this one

    "A sample of 24 independent observation is taken from a normal distribution of unknown mean μ but known variance σ2 = 96.42.
    The sample mean is 61.88 and sample variance is 57. What is the width of the 98% confidence interval for μ ?"
    7.17?
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    (Original post by Zacken)
    7.17?
    Apparently, 9.3255, but could be another mistake . Thanks for the help tho
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    (Original post by Mentally)
    Apparently, 9.3255, but could be another mistake . Thanks for the help tho
    Darn. Always sucks when the textbook answers are wrong.
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    (Original post by Mentally)
    Apparently, 9.3255, but could be another mistake . Thanks for the help tho
    9.3255 is the correct value.

    What's your calculation?

    Note: You need to use the critical value to 4 dec.pl. to get their exact answer.
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    (Original post by Mentally)
    "The time it takes for a cheese and ham sandwhich to be toasted in a Cass ground floor cafetaria appears to be normally distributed. A random sample of the times for toasting 20 such sandwhich had a sample mean of 151 seconds and a sample standard deviation of 21.9 seconds.


    What is the value of the upper end point of the 99% confidence interval for the mean time to toast a cheese and ham sandwich"


    I got 163.61 but apparently the answer is 165.0098, any idea how they got this?.
    Try the t-distribution - small sample. Book's answer is correct.
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    (Original post by ghostwalker)
    Try the t-distribution - small sample. Book's answer is correct.
    Thanks so much I got the answer now. But doesn't the central limit theorem say I should be able to approximate with a Normal Distribution, and if not then how large should sample size be before I can begin to estimate using Normal Distribution
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    (Original post by ghostwalker)
    9.3255 is the correct value.

    What's your calculation?

    Note: You need to use the critical value to 4 dec.pl. to get their exact answer.
    My answer is wayyy off, can you post working please?
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    (Original post by Mentally)
    Thanks so much I got the answer now. But doesn't the central limit theorem say I should be able to approximate with a Normal Distribution, and if not then how large should sample size be before I can begin to estimate using Normal Distribution
    Depends how accurate you want your results! If we focus attention on 99% confidence intervals, then the following bit of R code produces the attached graph.

    The red line is the value of the normal 0.995 quantile and the blue that of the t distribution as the number of degrees of freedom varies. At 19 degrees of freedom (the value in the problem above) the relative error is about 11%; at 100 degrees of freedom it is just under 2%. The relative error gets worse as you go further out into the tails of the distributions.
    Code:
    x <- 10:100 
    y <- qt(0.995, x) 
    plot(x,y, type="l", lwd=2, col="blue", xlab="degrees of freedom", ylab="", ylim=c(2.5,3.2)) 
    abline(h=qnorm(0.995), lwd=2, col="red" )
    Attached Images
     
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    (Original post by Mentally)
    But doesn't the central limit theorem say I should be able to approximate with a Normal Distribution, and if not then how large should sample size be before I can begin to estimate using Normal Distribution
    You should have some guidelines on how large a sample, either from your textbook or lectures. Crawshaw & Chambers suggests n>= 30 if you wish to approximate, but that's just an A-level text: As this is undergraduate level you may have different criteria.

    (Original post by Mentally)
    My answer is wayyy off, can you post working please?
    98% confidence interval implies 1% either end.
    So for your critical value, you're looking for z, such that \phi(z) = 1-0.01 = 0.99

    Then s.d. = \sqrt{\frac{96.42}{24}}

    Multiply by your z value and then double it for the interval length.
 
 
 
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