he level of nitrogen oxides (NOX) in the exhaust of a particular car model varies with mean 0.7 grams per mile and standard deviation 0.19 grams per mile .
(a) What sample size is needed so that the standard deviation of the sampling distribution is 0.01 grams per mile ?
Long form:
0.19 / (square root of n) = 0.01
multiply both sides by (square root of n) which gives you:
0.19 = (0.01) x (square root of n)
divide both sides by 0.01 which gives you:
0.19 / 0.01 = square root of n
both sides to the power of 2:
(0.19 / 0.01)^2 = (square root of n)^2
0.0361 / 0.0001= n
n = 361
Short form:
n = (0.19)^2 / (0.01)^2
n = 0.0361 / 0.0001
n = 361
Therefore the sample size that is needed is 361.
Well, finding the sample size needed is no small feat, but I'll give it a shot!
To determine the sample size needed, we can use the formula:
n = (Z * σ / E)^2,
where:
n = sample size
Z = Z-score (reliability coefficient)
σ = standard deviation of the population
E = desired margin of error
In this case, we want the standard deviation of the sampling distribution (E) to be 0.01 grams per mile. The population standard deviation (σ) is given as 0.19 grams per mile.
Now, we need to find the Z-score (reliability coefficient) corresponding to the desired level of confidence. Since that information is not provided, I'm going to assume a common confidence level of 95% (Z = 1.96).
Using these values, we can substitute them into the formula:
n = (1.96 * 0.19 / 0.01)^2
Calculating this, we get:
n ≈ 7246.
So, you would need a sample size of approximately 7246 to ensure that the standard deviation of the sampling distribution is 0.01 grams per mile.
Hope that brought a smile to your face!
To find the sample size needed so that the standard deviation of the sampling distribution is 0.01 grams per mile, you can use the formula:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score for the desired level of confidence (we'll assume a 95% confidence level, which corresponds to a Z-score of approximately 1.96)
σ = standard deviation of the population
E = desired margin of error (in this case, 0.01 grams per mile)
Plugging in the values:
n = (1.96 * 0.19 / 0.01)^2
Calculating this:
n = (0.3724 / 0.01)^2
n = 37.24^2
n ≈ 1384.7376
Rounding up to the nearest whole number, we get:
n ≈ 1385
Therefore, a sample size of approximately 1385 is needed to have a standard deviation of the sampling distribution of 0.01 grams per mile.
To find the sample size needed so that the standard deviation of the sampling distribution is 0.01 grams per mile, we can use the formula for the standard deviation of the sampling distribution:
\[ \text{Sampling Distribution Standard Deviation} = \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \]
In this case, the population standard deviation (σ) is given as 0.19 grams per mile, and we want the sampling distribution standard deviation to be 0.01 grams per mile. Let's denote the sample size as n and the desired sampling distribution standard deviation as SD.
Substituting the given values into the formula, we get:
\[ SD = \frac{0.19}{\sqrt{n}} \]
To solve for the sample size, we can rearrange the formula as follows:
\[ \sqrt{n} = \frac{0.19}{SD} \]
Now, let's solve for n:
\[ n = \left(\frac{0.19}{SD}\right)^2 \]
Substituting the given value for SD (0.01 grams per mile), we can calculate the sample size:
\[ n = \left(\frac{0.19}{0.01}\right)^2 \]
Simplifying the expression, we have:
\[ n = \left(\frac{19}{1}\right)^2 \]
\[ n = 361 \]
Therefore, a sample size of 361 is needed so that the standard deviation of the sampling distribution is 0.01 grams per mile.