# 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,"

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., $$2012: 761-766): 115, 181, 255, 418, 441, 461, 516, 739, 743, 789, 807, 865, 924, 983, 1025, 1062, 1063, 1165, 1191, 1222, 1222, 1251, 1277, 1290, 1357, 1369, 1408, 1455, 1278, 1519, 1578, 1578, 1599, 1603, 1605, 1696, 1735, 1799, 1815, 1852, 1899, 1925, 1965.$$
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 data exhibits a reasonably linear pattern.]
b) Calculate and interpret a confidence interval with a 99% confidence level for true average lifetime. [Hint: mean $$= 1191.6, s = 506.6$$.]

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doplovif
From the given information,
Sample data:
$${115, 181, 255, 418, 441, 461, 516, 739, 743, 789, 807, 865, 924, 983, 1025, 1062, 1063, 1165, 1191, 1222, 1222, 1251, 1277, 1290, 1357, 1369, 1408, 1455, 1278, 1519, 1578, 1578, 1599, 1603, 1605, 1696, 1735, 1799, 1815, 1852, 1899, 1925, 1965}$$
Since, sample size is greater than 30 but population standard deviation is not known therefore t-distribution will be most suitable here. (Though some statistician would use Z as well because by CLT, sample standard deviation can be approximated to population standard deviation)
b) The $$99\%$$ confidence interval for the average lifetime can be calculated as:
The mean and standard deviation can be computed as:
$$\displaystyle\overline{{x}}=\frac{{\sum{x}}}{{n}}$$
$$\displaystyle=\frac{{{115}+{181}+\ldots\ldots\ldots\ldots{.1965}}}{{43}}$$
$$\displaystyle={1186.279}$$
$$\displaystyle{s}=\sqrt{{\frac{{\sum{\left({x}_{{i}}-\overline{{x}}_{{i}}\right)}^{2}}}{{{n}-{1}}}}}$$
$$\displaystyle=\sqrt{{\frac{{{\left({115}-{1186.279}\right)}^{2}+\ldots\ldots\ldots\ldots+{\left({1965}-{1186.279}\right)}^{2}}}{{{43}-{1}}}}}$$
$$\displaystyle={506.1959}$$
CRITICAL VALUE:
Thus, confidence interval:
$$\displaystyle{C}{I}=\overline{{x}}\pm{t}_{{\alpha,{d}{f}}}\times\frac{s}{{\sqrt{{n}}}}$$
$$\displaystyle={1186.279}\pm{2.419}\times\frac{506.1959}{{\sqrt{{43}}}}$$
$$\displaystyle{\left({999.546},{1373.011}\right)}$$
Interpretation: This confidence interval shows that the $$99\%$$ of the times true population mean will fall in the interval or with repeated sampling the 0.99 probability that true average of lifetime will lie in the (999.546,1373.011).