Question

Many people feel that the drying of pavement marking paint is much too slow. You spend several days looking for the fastest drying paint you can find; you plan to measure the time (in seconds) for this paint to dry. From information provided by the paint supplier, you believe the time to dry is normally distributed with a standard deviation of 3 seconds. How many paint samples would you need to test to be able to obtain an estimate of paint drying time that is within 2 seconds of the actual mean drying time with a probability of 99.5%?

Answer 1

To estimate the mean drying time within 2 seconds of the actual mean drying time with a probability of 99.5%, you would need to test at least 5 paint samples.

Step-by-step explanation:

To determine the number of paint samples needed to estimate the mean drying time within 2 seconds of the actual mean drying time with a probability of 99.5%, we can use the formula for sample size:

n = (Z * σ) / E

Where:
- n represents the sample size
- Z is the Z-score corresponding to the desired probability (99.5%)
- σ is the standard deviation of the drying time (3 seconds in this case)
- E is the desired margin of error (2 seconds)

By plugging in the values, we can calculate the sample size:

n = (2.807 * 3) / 2 = 4.211

Since you cannot have a fraction of a sample, you would need to round up the sample size to 5. Therefore, you would need to test at least 5 paint samples to obtain an estimate of the paint drying time that is within 2 seconds of the actual mean drying time with a probability of 99.5%.

Answer 2

21

Step-by-step explanation:

Using Simple Random Sampling, we can estimate the sample size by the formula

`\bf n=(Z^2S^2)/(e^2)`

where

n = sample size

Z = the z-score corresponding to the confidence level 99.5%

S = the assumed standard deviation = 3 seconds

e = margin of error = 2 seconds

It is worth noticing that the higher the confidence level, the larger the sample should be.

The z-score corresponding to a confidence level of 99.5% can be obtained either with a table or the computer and equals

Z = 3.023

Replacing the values in our formula

`\bf n=((3.023)^23^2)/(2^2)=20.5616\approx 21`

So the size of the sample should be at least 21.