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# Stats ques. watch

1. "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?.
2. (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?
3. (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
4. (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 .
5. (Original post by B_9710)
What's the question.
So .
see op now
6. (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.
7. (Original post by Zacken)
I got 163.61 too and I'm inclined to agree with you.

"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 ?"
8. (Original post by Zacken)
I got 163.61 too and I'm inclined to agree with you.
Same.
9. (Original post by Mentally)

"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?
10. (Original post by Zacken)
7.17?
Apparently, 9.3255, but could be another mistake . Thanks for the help tho
11. (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.
12. (Original post by Mentally)
Apparently, 9.3255, but could be another mistake . Thanks for the help tho
9.3255 is the correct value.

Note: You need to use the critical value to 4 dec.pl. to get their exact answer.
13. (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.
14. (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
15. (Original post by ghostwalker)
9.3255 is the correct value.

Note: You need to use the critical value to 4 dec.pl. to get their exact answer.
16. (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

17. (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)
98% confidence interval implies 1% either end.
So for your critical value, you're looking for z, such that

Then s.d. =

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

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