Stats ques.

"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?.

(edited 8 years ago)

Reply 1

Zacken

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?

Reply 2

Mentally

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

Reply 3

B_9710

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)

(edited 8 years ago)

Reply 4

Mentally

Original post by B_9710

What's the question.
So

X~N(151, 21.9^2)

see op now

Reply 5

Zacken

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.

Reply 6

Mentally

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 ?"

Reply 7

B_9710

Original post by Zacken

I got 163.61 too and I'm inclined to agree with you.

Same.

Reply 8

Zacken

Original post by Mentally

7.17?

Reply 9

Mentally

Original post by Zacken

7.17?

Apparently, 9.3255, but could be another mistake . Thanks for the help tho

Reply 10

Zacken

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.

Reply 11

ghostwalker

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.

(edited 8 years ago)

Reply 12

ghostwalker

Original post by Mentally

Try the t-distribution - small sample. Book's answer is correct.

Reply 13

Mentally

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

Reply 14

Mentally

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?

Reply 15

Gregorius

Original post by Mentally

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.


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" )

(edited 8 years ago)

t-interval.png4.0KB

Reply 16

ghostwalker

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?