Confidence interval: mean working life of car components

[Monkeyseat]

Hi,

Question

A car manufacturer purchases large quantities of a particular component. The working lives of the components are known to be normally distributed with mean 2400 hours and standard deviation 650 hours. The manufacturer is concerned about the large variability and the supplier undertakes to improve the design so that the standard deviation is reduced to 300 hours.

A random sample of five of the new components is tested and found to last:

2730, 3120, 2980, 2680, 2800 hours.

Assuming that the lives of the new components are normally distributed with standard deviation 300 hours, calculate a 90% confidence interval for their mean working life.

Working

Mean = 2862

Therefore, the confidence interval for the mean is:

2862 +- (1.6449 * (300/sqrt. 5)) = 2641 hours to 3083 hours (to the nearest hour)

The book says the answer is 2640 hours to 3080 hours. I just wanted to check whether I have made a mistake or the book has made a typo/rounded it strangely.

Thanks.

 

[chivox]

Just a reminder: All the hours in your sample have three significant digits. It may not be necessary to compute the value to the nearest hour, only to the nearest 10 hours, since three significant digits is all you used in your calculations anyway.

 

[Monkeyseat]

I see what you mean chivox. Is what I have done correct though?

Thanks.

 

[Monkeyseat]

Update

Hi, I just have another question.

"Is there any reason to doubt the assumption that the standard deviation of the lives of the new components is 300 hours?"

The answer according to the book is: "Some doubt as if standard deviation is 300, the sample range is only about 1.5 standard deviations (or estimated standard deviation only 184)."

Would someone be able to explain this? First, what is the sample range? Is it the highest number in the sample minus the lowest number in the sample (3120 - 2680 = 440) or is it the upper limit of the confidence interval minus the lower limit of the confidence interval (3083 - 2641 = 442)?

Second, I assume 440 is around 1.5 standard deviations because 300 * 1.5 = 450, but why does it matter that "the sample range is only about 1.5 standard deviations"? Is it to do with a certain amount of data being within a certain amount of standard deviations (i.e. 99.7% of area lies in the range: mean +- 3 standard deviations)?

Thanks!

EDIT:

Correction, estimated standard deviation only 184.

 

[Deleted member 4993]

Re: Update

Hi, I just have another question.

"Is there any reason to doubt the assumption that the standard deviation of the lives of the new components is 300 hours?"

The answer according to the book is: "Some doubt as if standard deviation is 300, the sample range is only about 1.5 standard deviations (or estimated standard deviation only 189)."

Would someone be able to explain this? First, what is the sample range? Is it the highest number in the sample minus the lowest number in the sample (3120 - 2680 = 440) correct or is it the upper limit of the confidence interval minus the lower limit of the confidence interval (3083 - 2641 = 442)?

Second, I assume 440 is around 1.5 standard deviations because 300 * 1.5 = 450, but why does it matter that "the sample range is only about 1.5 standard deviations"? Is it to do with a certain amount of data being within a certain amount of standard deviations (i.e. 99.7% of area lies in the range: mean +- 3 standard deviations)?correct - check in z -distribution table - how much area is covered under ±0.75 range

Thanks!

 

[Monkeyseat]

Re: Update

correct - check in z -distribution table - how much area is covered under ±0.75 range[/color]

So:

P(-0.75 < Z < 0.75) = P(Z < 0.75) - P(Z < -0.75)
P(-0.75 < Z < 0.75) = 0.77337 - (1 - 0.77337)
P(-0.75 < Z < 0.75) = 0.54674

Is that correct? If so, what does it show? Not all of the data is included? Sorry to keep asking questions.

Thanks.

 

[Deleted member 4993]

Re: Update

correct - check in z -distribution table - how much area is covered under ±0.75 range[/color]

So:

P(-0.75 < Z < 0.75) = P(Z < 0.75) - P(Z < -0.75)
P(-0.75 < Z < 0.75) = 0.77337 - (1 - 0.77337)
P(-0.75 < Z < 0.75) = 0.54674

Is that correct? If so, what does it show? Sorry to keep asking questions. Not all of the data is included?

There is about 55% chance that a data point randomly selected from this sample space will belong to the population. That makes the confidence level on the calculated sample standard deviation very low.

Thanks.

 

[Monkeyseat]

Hi, there are a few basic things I'm unsure about.

1) Would it be correct to say:

"Because the sample range is around 1.5 standard deviations, all of the data from the sample lies within 0.75 standard deviations of the sample mean"? I don't know whether I am right here - is it the sample or population? And is it 0.75 standard deviations or 1.5 standard deviations (I think it's 0.75)?

2) Why do you calculate the area covered under the range +-0.75, instead of +-1.5?

3) What did the answer in the book mean when it said "estimated standard deviation only 184"? Where did this come from?

4) If you said 95.5% of data lies within the range "mean +- 2 standard deviations", would the z values be +-2? And z = +-3 for "mean +-3 standard deviations" etc.?

5) If the total area under the curve = 1, is it therefore correct that P(z<0) = 0.5 (seeing as the tables do not give a z value for 0)?

I apologise for asking so many questions - I think they are quick to answer. Please excuse my ignorance, I have not studied this topic very long. I am still trying to get my head around it.

Thanks.