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Analysis showed that the mean arrival rate for vehicles at a certain Shell station on Friday afternoon last year was 4.5 vehicles per minute. How large a sample would be needed to estimate this year's mean arrival rate with 98 percent confidence and an error of ± 0.5? (Round the standard deviation answer to 4 decimal places. Enter your answer as a whole number (no decimals). Use a z-value taken to four decimal places in your calculations.) Sample size

Respuesta :

Answer:

The sample size needed is n=25.

Step-by-step explanation:

We will use a Poisson process to model the arrival.

We know the mean rate of arrivals, that is

[tex]\lambda=4.5[/tex]

The standard deviation is calculated as:

[tex]\sigma=\sqrt{\lambda}=\sqrt{4.5}=2.1213[/tex]

The z-value for a 98% CI is z=2.3262.

If the 98% CI has to be within a error of 0.5, we have:

[tex]UL-LL=2z\sigma/\sqrt{n}=2*0.5=1\\\\\sqrt{n}=z\sigma=2.3262*2.1213=4.9346\\\\n=4.9346^2=24.35\approx25[/tex]

The sample size needed is n=25.

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