Question

4.5.t Data from an AASHO road test revealed that the deviations from the planned thickness of pavement layers were very nearly normally dis- tributed with . = 0.195 in. ž =-0.02 in, The sample size was 4217. (a) What are the 95 percent confidence limits on m assuming (i)o 0.195 in. (ii) ơ is not known with certainty. (b) Adopting the normal model using f and s as point estimates of m and ơ, what is the probability that a particular specimen will exceed the tolerances- 0.02 ft and +o.04 ft?

Answer 1

This is a simple problem related to ascertaining the confidence interval of the mean when the population variance in known and when it is not known .Further we also need to estimate the probability that the mean lies in a particular domain .

1). We shall find the confidence interval firstly when the population standard deviation is provided.

The provided sample mean is X =-0.02 and the population standard deviation is σ = 0.195. The size of the sample is n = 4217 and the required confidence level is 95%. Based on the provided information, the critical z-value for α-0.05 is ~c-1.96. The 95% confidence for the population mean μ is computed using the following expression Tm Therefore, based on the information provided, the 95 % confidence for the population mean is 1.96 x 0.195 1.96 x 0.195 V4217 4217 -(-0.02-0.006,-0.02+0.006) (-0.026,-0.014) which completes the calculation.

ii) . We shall now find the confidence interval firstly when the population standard deviation is not provided.

It has to be noted here that the sample standard deviation is provided instead of the population standard deviation. In cases like these we treat the sample standard deviation as an estimate of the population standard deviation.

The provided sample mean is X-_0.02 and the sample standard deviation is s = 0.195. The size of the sample is n 4217 and the required confidence level is 95%. The number of degrees of freedom are df = 4217-1 = 4216, and the significance level is α = 0.05 Based on the provided information, the critical t-value for α-0.05 and df-4216 degrees of freedom is te = 1.961. The 95% confidence for the population mean μ is computed using the following expression tc x s 7n Therefore, based on the information provided, the 95 % confidence for the population mean is cr =(-0.02-1961 217195,-002 + 13 1.961 × 0.195 1.961 x 0.195 CI = -0.02- V4217 = (-0.02-0.006,-0.02 0.006) = (-0.026,-0.014)

2). Now we are to find a probability of the a specimen exceeding the confidence interval of the point estimate of the sample mean when we take the sample standard deviation .

WE have seen that range of the confidence interval in case when we consider the mean and the sample standard deviation for it, is equal to

(-0.014-(-0.026)

=0.012

Now suppose a sample specimen exceeds the tolerance limits of -0.020

Then it can go upto a aminimum of -0.026 only (lower boundary of the confidence interval)

Thus it will have a -0.020 -(-0.026) =0.006 points to play with .

This will be our favourable cases.

Now when the limit exceeds +0.04 then it will not lie in the caclulated confidence limts since its upper cut off limit is -0.014

Thus there cant be any region within the confidence boundary to play with.

Hence favourable cases =0

Thus overall favourable cases =0.006+0 =0.006

Total number of allowable cases =range of the confidence interval

=0.012

Thus probability =favourable cases/total cases

=0.006/0.012

=50 %

Thus there is a 50 % probability that the tolerance limit of -0.02 and 0.04 ft will be exceeded.