Question

The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l. a) If displaystyle{X}_{{1}},{

Confidence intervals
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asked 2021-03-07
The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l.
a) If \(\displaystyle{X}_{{1}},{X}_{{2}},\ldots,{X}_{{n}}\) are the times, in minutes, between successive customers selected randomly, estimate the parameter of the distribution.
b) The randomly selected 12 times between successive customers are found as \(\displaystyle{1.8},{1.2},{0.8},{1.4},{1.2},{0.9},{0.6},{1.2},{1.2},{0.8},{1.5},{\quad\text{and}\quad}{0.6}\) mins. Estimate the mean time between successive customers, and write down the distribution function.
c) In order to estimate the distribution parameter with 0.3 error and \(4\%\) risk, find the minimum sample size.

Answers (1)

2021-03-08
Given
The time between successive customers coming to the market X is assumed to have Exponential distribution with parameter I.
\(\displaystyle{X}\approx{E}{x}{p}{\left({I}\right)}\)
a)
The mean of exponential distribution with parameter I is I.
As the sample mean is unbiased estimate of population mean we say the following
\(\displaystyle\frac{1}{\hat{{I}}}=\frac{{\sum{x}_{{i}}}}{{n}}\)
This is the estimate for the parameter of the distribution.
b)
The data given 12 randomly selected times between successive customers as shown below
\(\displaystyle{1.8},{1.2},{0.8},{1.4},{1.2},{0.9},{0.6},{1.2},{1.2},{0.8},{1.5},{0.6}\)
The parameter is estimated as shown below
\(\displaystyle\frac{1}{\hat{{I}}}=\frac{{\sum{x}_{{i}}}}{{n}}\)
\(\displaystyle\frac{1}{\hat{{I}}}=\frac{{{1.8},{1.2},{0.8},{1.4},{1.2},{0.9},{0.6},{1.2},{1.2},{0.8},{1.5},{0.6}}}{{12}}\)
\(\displaystyle\hat{{I}}=\frac{10}{{11}}={0.9091}\)
So the exponential distribution is shown below
\(\displaystyle{P}{\left({X}={x}\right)}={I}{e}^{{{I}{x}}}{x}\ge{0}\)
c)
Given
The margin of error \(\displaystyle{E}={0.3}\)
Level of significance \(\displaystyle={4}\%\)
If X is exponential distribution with parameter I then X is approximately normal distribution with mean and variance \(\displaystyle\frac{1}{{I}}{\quad\text{and}\quad}{\left(\frac{1}{{I}}\right)}^{2}\) respectively.
We can approximate the distribution is normal with mean and variance as \(\displaystyle\frac{10}{{11}},{\left(\frac{10}{{11}}\right)}^{2}\) respectively.
Thus the Z critical score for \(4\%\) level of significance or \(96\%\) confidence interval is
\(\displaystyle{Z}_{{{1}-{0.04}\text{/}{2}}}={Z}_{{0.98}}={2.05}\)
The sample size required is calculated as shown below
\(\displaystyle{E}=\frac{{{Z}\sigma}}{\sqrt{{n}}}\)
\(\displaystyle{n}={\left(\frac{{{Z}\sigma}}{{E}}\right)}^{2}\)
\(\displaystyle{n}={\left(\frac{{\sqrt[{{2.05}}]{{\frac{10}{{11}}}}}}{{0.3}}\right)}^{2}\)
\(\displaystyle{n}={42.45}\)
Thus we can say the minimum sample size required is 43. (rounding to next integer).
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