a) If

b) The randomly selected 12 times between successive customers are found as

c) In order to estimate the distribution parameter with 0.3 error and

Tobias Ali
2021-03-07
Answered

The time between successive customers coming to the market is assumed to have Exponential distribution with parameter l.

a) If$X}_{1},{X}_{2},\dots ,{X}_{n$ 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$1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}0.6$ 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$4\mathrm{\%}$ risk, find the minimum sample size.

a) If

b) The randomly selected 12 times between successive customers are found as

c) In order to estimate the distribution parameter with 0.3 error and

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Clelioo

Answered 2021-03-08
Author has **88** answers

Given

The time between successive customers coming to the market X is assumed to have Exponential distribution with parameter I.

$X\approx Exp\left(I\right)$

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

$\frac{1}{\hat{I}}=\frac{\sum {x}_{i}}{n}$

This is the estimate for the parameter of the distribution.

b)

The data given 12 randomly selected times between successive customers as shown below

$1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,0.6$

The parameter is estimated as shown below

$\frac{1}{\hat{I}}=\frac{\sum {x}_{i}}{n}$

$\frac{1}{\hat{I}}=\frac{1.8,1.2,0.8,1.4,1.2,0.9,0.6,1.2,1.2,0.8,1.5,0.6}{12}$

$\hat{I}=\frac{10}{11}=0.9091$

So the exponential distribution is shown below

$P(X=x)=I{e}^{Ix}x\ge 0$

c)

Given

The margin of error$E=0.3$

Level of significance$=4\mathrm{\%}$

If X is exponential distribution with parameter I then X is approximately normal distribution with mean and variance$\frac{1}{I}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{\left(\frac{1}{I}\right)}^{2}$ respectively.

We can approximate the distribution is normal with mean and variance as$\frac{10}{11},{\left(\frac{10}{11}\right)}^{2}$ respectively.

Thus the Z critical score for$4\mathrm{\%}$ level of significance or $96\mathrm{\%}$ confidence interval is

${Z}_{1-0.04\text{/}2}={Z}_{0.98}=2.05$

The sample size required is calculated as shown below

$E=\frac{Z\sigma}{\sqrt{n}}$

$n={\left(\frac{Z\sigma}{E}\right)}^{2}$

$n={\left(\frac{\sqrt[2.05]{\frac{10}{11}}}{0.3}\right)}^{2}$

$n=42.45$

Thus we can say the minimum sample size required is 43. (rounding to next integer).

The time between successive customers coming to the market X is assumed to have Exponential distribution with parameter I.

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

This is the estimate for the parameter of the distribution.

b)

The data given 12 randomly selected times between successive customers as shown below

The parameter is estimated as shown below

So the exponential distribution is shown below

c)

Given

The margin of error

Level of significance

If X is exponential distribution with parameter I then X is approximately normal distribution with mean and variance

We can approximate the distribution is normal with mean and variance as

Thus the Z critical score for

The sample size required is calculated as shown below

Thus we can say the minimum sample size required is 43. (rounding to next integer).

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