Question

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer (“A Goodness of Fit Approach to the Class of Life Distributions with Unknown Age,” Quality and Reliability Engr. Intl., \({\rm{2012: 761--766):}}\)

\(\begin{array}{*{20}{l}}{{\rm{115 181 255 418 441 461 516 739 743 789 807}}}\\{{\rm{865 924 983 1025 1062 1063 1165 1191 1222 1222 1251}}}\\{{\rm{1277 1290 1357 1369 1408 1455 1478 1519 1578 1578 1599}}}\\{{\rm{1603 1605 1696 1735 1799 1815 1852 1899 1925 1965}}}\end{array}\)

a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. (Note: A normal probability plot of the data exhibits a reasonably linear pattern.)

b. Calculate and interpret a confidence interval with a \({\rm{99\% }}\)confidence level for true average lifetime.

Short answer

a) Confidence interval can be calculated without assuming anything

b) The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E = 1191}}{\rm{.6279 - 198}}{\rm{.9474 = 992}}{\rm{.6805}}\\{\rm{\bar x + E = 1191}}{\rm{.6279 + 198}}{\rm{.9474 = 1390}}{\rm{.5753}}\end{array}\)

Step-by-step solution

Step 1: Given by

Given:

\({\rm{n = 43}}\)

\(\begin{array}{l}{\rm{115,181,255,418,441,461,516,739,743,789,807,865,924,983,1025,1062,1063,1}}\\{\rm{165,1191,1222,1222,1251,1277,1290,1357,1369,1408,1455,1478,1519,1578,1578,1599,1603,}}\\{\rm{ 1605,1696,1735,1799,1815,1852,1899,1925,1965}}\end{array}\)

Central limit theorem: If the sample size is large ,then the sampling distribution of the sample mean \({\rm{\bar x}}\)is approximately normal.

The data set contains 43 data values, thus we can use the central limit theorem to determine the shape of the distribution, because we then know that the average lifetime has approximately a normal distribution.

To plot Normal probability

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.

If the data contains n data values, then the standardized normal scores are the z-scores in the normal probability table of the appendix corresponding to an area of \(\frac{{{\rm{j - 0}}{\rm{.5}}}}{{\rm{n}}}\) (or the closest area) with \({\text{j^I \{ 1,2,3, ldots ,n\} }}\).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

The pattern in the normal probability plot is roughly linear, which indicate that the population distribution of x is approximately normal.

Since the distribution of x is approximately normal, the sampling distribution of the sample mean \({\rm{\bar x}}\)is also approximately normal.

Since the sampling distribution of the sample mean \({\rm{\bar x}}\)is approximately normal by the central limit theorem, we do not need to assume that the population distribution is approximately normal.

Hence confidence interval can be calculated without assuming anything.

To Calculate and interpret a confidence interval

(b) Given:

\({\rm{n = 43\bar x = 1191}}{\rm{.6s = 506}}{\rm{.6c = 99\% = 0}}{\rm{.99}}\)

\(\begin{array}{l}{\rm{115,181,255,418,441,461,516,739,743,789,807,865,924,983,}}\\{\rm{1025,1062,1063,1165,1191,1222,1222,1251,1277,1290,1357,1369,}}\\{\rm{1408,1455,1478,1519,1578,1578,1599,1603,}}\\{\rm{1605,1696,1735,1799,1815,1852,1899,1925,1965}}\end{array}\)

Central limit theorem: If the sample size is large , then the sampling distribution of the sample mean \({\rm{\bar x}}\)is approximately normal.

The data set contains 43 data values, thus we can use the central limit theorem to determine the shape of the distribution and we thus know that the sampling distribution of the sample mean is approximately normal.

Large-sample confidence interval

The sample mean is the sum of all data values divided by the sample size:

\({\rm{\bar x = }}\frac{{{\rm{115 + 181 + 255 + \ldots + 1899 + 1925 + 1965}}}}{{{\rm{43}}}}{\rm{ = }}\frac{{{\rm{51240}}}}{{{\rm{43}}}}{\rm{\gg 1191}}{\rm{.6279}}\)

The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance:

\({\rm{s = }}\sqrt {\frac{{{{{\rm{(115 - 1191}}{\rm{.6279)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (1965 - 1191}}{\rm{.6279}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{43 - 1}}}}} {\rm{\gg 506}}{\rm{.6350}}\)

For confidence level \({\rm{1 - \alpha = 0}}{\rm{.99}}\), determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\)using the normal probability table in the appendix (look up 0.005 in the table, the z-score is then the found z score with opposite sign):

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.575}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.575 \times }}\frac{{{\rm{506}}{\rm{.6350}}}}{{\sqrt {{\rm{43}}} }}{\rm{\gg 198}}{\rm{.9474}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E = 1191}}{\rm{.6279 - 198}}{\rm{.9474 = 992}}{\rm{.6805}}\\{\rm{\bar x + E = 1191}}{\rm{.6279 + 198}}{\rm{.9474 = 1390}}{\rm{.5753}}\end{array}\)