Confidence interval: maximum load for stud anchors

[Monkeyseat]

Hi,

Question

Stud anchors are used in the construction industry. Samples are tested by embedding them in concrete and applying a steadily increasing load until the anchor fails. A sample of six tests gave the following maximum loads in kilonewtons:

27.0, 30.5, 28.0, 23.0, 27.5, 26.5

a) Assuming a normal distribution for the maximum load calculate a 95% confidence interval for the mean.
b) If the mean was at the lower end of the interval calculated in (a) estimate the value, k, which the maximum load would exceed with probability 0.99. Assume the standard deviation estimated is an accurate assessment of the population standard deviation.

Safety regulations require that the greatest load which may be applied under working conditions is ("x bar" - 2s)/3, where "x bar" and s are calculated from a sample of six tests.
c) Calculate this value and comment on the adequacy of this regulation in these circumstances.

Working

a) 24.52-29.64

b) I'm not sure how to do this. I tried this method but I'm not sure whether it gave the correct answer:

z = (k - mean)/standard deviation
-2.3263 = (k - 24.52)/2.438
-5.67... = k - 24.52
k = 18.85

Is that the correct method/answer? If the method is correct, I don't know whether the z value should be 2.3263 or -2.3263, so could someone say which is correct and why?

c) I calculated the value and got 7.40, but I don't really know what comment to make (I think the comment depends on my answer in part (b) but I don't know whether that answer is correct)...

Any help with part (b) and (c) would be much appreciated.

Thanks.

 

[Deleted member 4993]

Hi,

Question

Stud anchors are used in the construction industry. Samples are tested by embedding them in concrete and applying a steadily increasing load until the anchor fails. A sample of six tests gave the following maximum loads in kilonewtons:

27.0, 30.5, 28.0, 23.0, 27.5, 26.5

a) Assuming a normal distribution for the maximum load calculate a 95% confidence interval for the mean.
b) If the mean was at the lower end of the interval calculated in (a) estimate the value, k, which the maximum load would exceed with probability 0.99. Assume the standard deviation estimated is an accurate assessment of the population standard deviation.

Safety regulations require that the greatest load which may be applied under working conditions is ("x bar" - 2s)/3, where "x bar" and s are calculated from a sample of six tests.
c) Calculate this value and comment on the adequacy of this regulation in these circumstances.

Working

a) 24.52-29.64

b) I'm not sure how to do this. I tried this method but I'm not sure whether it gave the correct answer:

z = (k - mean)/standard deviation
-2.3263 = (k - 24.52)/2.438
-5.67... = k - 24.52
k = 18.85

Is that the correct method/answer? If the method is correct, I don't know whether the z value should be 2.3263 or -2.3263, so could someone say which is correct and why?

c) I calculated the value and got 7.40, but I don't really know what comment to make (I think the comment depends on my answer in part (b) but I don't know whether that answer is correct)...

Any help with part (b) and (c) would be much appreciated.

Thanks.

Before we start anything -

calculate the sample mean and the sample standard deviation of the given set of data

Then tell us how would you answer (a) from those two values (and number of observation).

got to:

http://www.stat.yale.edu/Courses/1997-9 ... onfint.htm

for some worked examples.

 

[Monkeyseat]

Subhotosh Khan, I had no problem with part (a). I had already calculated the sample standard deviation and sample mean, and calculated the confidence interval. I didn't bother posting any working because I didn't need any help with that part and knew the answer was correct.

Sample mean = 27.083. Sample standard deviation = 2.438.

I did: 27.083 +- 2.571 * (2.438/(sqrt. 6)) = 24.52 to 29.64

I mainly just need help with part (b), specifically (if my method was correct) which z value to use (2.3263 or -2.3263) and why.

Thanks.